An α-particle of energy 5MeV is scattered through 180° by gold nucleus. The distance of closest approach is of the order of
🥳 Wohoo! Correct answer
K.E. = kq₁q₂/r
Use K.E. = (kq₁q₂)/r at closest approach
Convert 5MeV to joules and substitute charges
Calculate r ≈ 10⁻¹² cm
Wrong energy conversion
😢 Uh oh! Incorrect answer, Try again
Consider Coulomb potential
K.E. = kq₁q₂/r
Use K.E. = (kq₁q₂)/r at closest approach
Convert 5MeV to joules and substitute charges
Calculate r ≈ 10⁻¹² cm
Wrong energy conversion
A radioactive decay can form an isotope of the original nucleus with the emission of particles
🥳 Wohoo! Correct answer
ΔZ, ΔA rules
α particle changes mass by -4, atomic number by -2 β⁻ particle changes mass by 0, atomic number by +1
Track changes in mass number and atomic number
One α and two β gives stable isotope
Wrong particle counting
😢 Uh oh! Incorrect answer, Try again
Balance nuclear equation
ΔZ, ΔA rules
α particle changes mass by -4, atomic number by -2 β⁻ particle changes mass by 0, atomic number by +1
Track changes in mass number and atomic number
One α and two β gives stable isotope
Wrong particle counting
A nucleus at rest splits into two nuclear parts having radii in the ratio 1:2. Their velocities are in the ratio
🥳 Wohoo! Correct answer
m ∝ r³
Use conservation of momentum: m₁v₁ = m₂v₂ Mass ratio m₁:m₂ = r₁³:r₂³ = 1:8
v₁/v₂ = m₂/m₁ = 8/1
Therefore velocity ratio is 8:1
Forgetting cubic relation
😢 Uh oh! Incorrect answer, Try again
Consider mass-radius relation
m ∝ r³
Use conservation of momentum: m₁v₁ = m₂v₂ Mass ratio m₁:m₂ = r₁³:r₂³ = 1:8
v₁/v₂ = m₂/m₁ = 8/1
Therefore velocity ratio is 8:1
Forgetting cubic relation
The half-life of a radioactive substance is 20 minutes. The time taken between 50% decay and 87.5% decay of the substance will be
🥳 Wohoo! Correct answer
N = N₀e^(-λt)
After first half-life: 50% remains Initial activity N₀/2
Final activity = 12.5% = N₀/8 Use N = N₀(1/2)^(t/T₁/₂)
Solve to get t = 40 minutes
Wrong time calculation
😢 Uh oh! Incorrect answer, Try again
Use decay equation
N = N₀e^(-λt)
After first half-life: 50% remains Initial activity N₀/2
Final activity = 12.5% = N₀/8 Use N = N₀(1/2)^(t/T₁/₂)
Solve to get t = 40 minutes
Wrong time calculation
What is the wavelength of light for the least energetic photon emitted in the Lyman series of the hydrogen spectrum (take hc = 1240 eV nm)
🥳 Wohoo! Correct answer
λ = hc/ΔE
For Lyman series, transitions end at n₁=1 Least energetic means starting from n₂=2
Use λ = 1/R(1/n₁² - 1/n₂²) = 1/0.0097(1/1² - 1/2²)
Calculate to get λ = 122 nm
Wrong series identification
😢 Uh oh! Incorrect answer, Try again
Consider least energy transition
λ = hc/ΔE
For Lyman series, transitions end at n₁=1 Least energetic means starting from n₂=2
Use λ = 1/R(1/n₁² - 1/n₂²) = 1/0.0097(1/1² - 1/2²)
Calculate to get λ = 122 nm
Wrong series identification
If an electron in a hydrogen atom jumps from an orbit of level n=3 to an orbit of level n=2, the emitted radiation has a frequency (R=Rydberg constant, C=velocity of light)
🥳 Wohoo! Correct answer
ΔE = -13.6(1/n₁² - 1/n₂²)
Use ΔE = hν = (13.6 eV)(1/n₁² - 1/n₂²)
Substitute n₁=2, n₂=3
Simplify to get ν = 5RC/36
Wrong energy calculation
😢 Uh oh! Incorrect answer, Try again
Use energy level difference
ΔE = -13.6(1/n₁² - 1/n₂²)
Use ΔE = hν = (13.6 eV)(1/n₁² - 1/n₂²)
Substitute n₁=2, n₂=3
Simplify to get ν = 5RC/36
Wrong energy calculation
Find the de-Broglie wavelength of an electron with a kinetic energy of 120eV
🥳 Wohoo! Correct answer
λ = h/p = h/√(2mKE)
Convert KE to joules: 120×1.6×10⁻¹⁹ J
Calculate momentum: p = √(2mKE)
λ = h/p gives 112 pm
Unit conversion errors
😢 Uh oh! Incorrect answer, Try again
Use wave-particle duality
λ = h/p = h/√(2mKE)
Convert KE to joules: 120×1.6×10⁻¹⁹ J
Calculate momentum: p = √(2mKE)
λ = h/p gives 112 pm
Unit conversion errors
Light of two different frequencies whose photons have energies 1eV and 2.5eV respectively, successively illuminate a metallic surface whose work function is 0.5eV. Ratio of maximum speeds of emitted electrons will be
🥳 Wohoo! Correct answer
KE = hν - φ
KE₁ = hν₁ - φ = 1 - 0.5 = 0.5 eV
KE₂ = hν₂ - φ = 2.5 - 0.5 = 2 eV
Since KE = ½mv², v₁:v₂ = √(0.5):√2 = 1:2
Not taking square root
😢 Uh oh! Incorrect answer, Try again
Consider photoelectric equation
KE = hν - φ
KE₁ = hν₁ - φ = 1 - 0.5 = 0.5 eV
KE₂ = hν₂ - φ = 2.5 - 0.5 = 2 eV
Since KE = ½mv², v₁:v₂ = √(0.5):√2 = 1:2
Not taking square root
A nucleus of mass 20u emits a γ photon of energy 6 MeV. If emission occurs when nucleus is free and at rest, nucleus will have kinetic energy nearest to (1u = 1.6×10⁻²⁷ kg)
🥳 Wohoo! Correct answer
E = p²/2M
Apply conservation of momentum: pᵧ = P
Calculate energy using E = p²/2M
Convert units and solve for kinetic energy = 1 KeV
Students might ignore recoil
😢 Uh oh! Incorrect answer, Try again
Consider recoil momentum
E = p²/2M
Apply conservation of momentum: pᵧ = P
Calculate energy using E = p²/2M
Convert units and solve for kinetic energy = 1 KeV
Students might ignore recoil
A radioactive sample of half-life 10 days contains 1000x nuclei. Number of original nuclei present after 5 days is
🥳 Wohoo! Correct answer
N = N₀e^(-λt)
Use decay formula: N = N₀e^(-λt)
Calculate λ = ln2/T½ = ln2/10
Solve: N = 1000x×e^(-ln2×5/10) = 707x
Students might divide by 2 directly
😢 Uh oh! Incorrect answer, Try again
Consider half time fraction
N = N₀e^(-λt)
Use decay formula: N = N₀e^(-λt)
Calculate λ = ln2/T½ = ln2/10
Solve: N = 1000x×e^(-ln2×5/10) = 707x
Students might divide by 2 directly
When an electron jumps from n = 4 level to n = 1 level, the angular momentum of electron changes by
🥳 Wohoo! Correct answer
L = nh/2π
Use quantization rule: mvr = nh/2π
Calculate difference: Δ(mvr) = (4-1)h/2π
Change in angular momentum = 3h/2π
Students might add quantum numbers
😢 Uh oh! Incorrect answer, Try again
Consider quantization of angular momentum
L = nh/2π
Use quantization rule: mvr = nh/2π
Calculate difference: Δ(mvr) = (4-1)h/2π
Change in angular momentum = 3h/2π
Students might add quantum numbers
Total energy of an electron in an excited state of a hydrogen atom is -3.4 eV. The kinetic and potential energy of an electron in this state
🥳 Wohoo! Correct answer
K = -E, U = 2E
Apply virial theorem: K = -E, U = 2E
Given E = -3.4 eV
Calculate: K = 3.4 eV, U = -6.8 eV
Students might ignore negative sign
😢 Uh oh! Incorrect answer, Try again
Use energy relationships in atom
K = -E, U = 2E
Apply virial theorem: K = -E, U = 2E
Given E = -3.4 eV
Calculate: K = 3.4 eV, U = -6.8 eV
Students might ignore negative sign
The de Broglie wavelength of an electron accelerated to a potential of 400 V is approximately
🥳 Wohoo! Correct answer
λ = h/√(2meV)
Use de Broglie formula: λ = h/p = h/√(2meV)
Substitute values in λ = 1.23/√V nm
Calculate: λ = 1.23/√400 = 0.06 nm
Students might forget to convert units
😢 Uh oh! Incorrect answer, Try again
Use simplified formula for electron wavelength
λ = h/√(2meV)
Use de Broglie formula: λ = h/p = h/√(2meV)
Substitute values in λ = 1.23/√V nm
Calculate: λ = 1.23/√400 = 0.06 nm
Students might forget to convert units
The mass defect of ⁴He is 0.03μ. The binding energy per nucleon of helium (in MeV) is
🥳 Wohoo! Correct answer
E = mc²
Convert mass defect to energy: E = mc² = 0.03 × 931 MeV
Total binding energy = 27.93 MeV
Per nucleon = 27.93/4 = 6.9825 MeV
Not dividing by number of nucleons
😢 Uh oh! Incorrect answer, Try again
Remember mass-energy conversion
E = mc²
Convert mass defect to energy: E = mc² = 0.03 × 931 MeV
Total binding energy = 27.93 MeV
Per nucleon = 27.93/4 = 6.9825 MeV
Not dividing by number of nucleons
The particles emitted in the decay of ²³⁸U to ²³⁴U
🥳 Wohoo! Correct answer
ΔA = 4 for α, ΔZ = 1 for β
Write nuclear equation
Check charge and mass conservation
Requires one alpha and two beta emissions
Not balancing nuclear equations
😢 Uh oh! Incorrect answer, Try again
Follow conservation laws
ΔA = 4 for α, ΔZ = 1 for β
Write nuclear equation
Check charge and mass conservation
Requires one alpha and two beta emissions
Not balancing nuclear equations
The scientist who is credited with the discovery of 'nucleus' in an atom is
🥳 Wohoo! Correct answer
No specific formula
Consider alpha particle scattering experiment
Rutherford observed large angle scattering
This led to discovery of nuclear model
Confusing different atomic models
😢 Uh oh! Incorrect answer, Try again
Alpha particle scattering was key experiment
No specific formula
Consider alpha particle scattering experiment
Rutherford observed large angle scattering
This led to discovery of nuclear model
Confusing different atomic models
The energy (in eV) required to excite an electron from n = 2 to n = 4 state in the hydrogen atom is
🥳 Wohoo! Correct answer
ΔE = E_final - E_initial
Use energy formula E = -13.6/n² eV
Calculate energies: E₂ = -13.6/4 eV, E₄ = -13.6/16 eV
Energy required = E₄ - E₂ = -13.6/16 - (-13.6/4) = +2.55 eV
Not considering sign conventions
😢 Uh oh! Incorrect answer, Try again
Energy is positive for excitation
ΔE = E_final - E_initial
Use energy formula E = -13.6/n² eV
Calculate energies: E₂ = -13.6/4 eV, E₄ = -13.6/16 eV
Energy required = E₄ - E₂ = -13.6/16 - (-13.6/4) = +2.55 eV
Not considering sign conventions
In a nuclear reactor the function of the Moderator is to decrease
🥳 Wohoo! Correct answer
KE = ½mv²
Consider role of moderator
Moderator slows fast neutrons to thermal energies
Slower neutrons have higher fission cross-section
Confusing moderator with control rods
😢 Uh oh! Incorrect answer, Try again
Fast neutrons must be slowed for fission
KE = ½mv²
Consider role of moderator
Moderator slows fast neutrons to thermal energies
Slower neutrons have higher fission cross-section
Confusing moderator with control rods
A proton and an α particle are accelerated through the same potential difference V. The ratio of their de Broglie wavelengths is
🥳 Wohoo! Correct answer
λ = h/√(2mqV)
Use de Broglie relation: λ = h/mv
Calculate v using K.E = ½mv² = qV
Compare wavelengths considering mass and charge differences
Not considering charge difference
😢 Uh oh! Incorrect answer, Try again
Consider mass and charge ratio of α particle
λ = h/√(2mqV)
Use de Broglie relation: λ = h/mv
Calculate v using K.E = ½mv² = qV
Compare wavelengths considering mass and charge differences
Not considering charge difference
The half-life of tritium is 12.5 years. What mass of tritium of initial mass 64 mg will remain undecayed after 50 years?
🥳 Wohoo! Correct answer
m = m₀(1/2)ⁿ
Calculate number of half-lives: 50/12.5 = 4
Use formula: m = m₀(1/2)ⁿ where n=4
Calculate: 64 × (1/16) = 4 mg
Not counting half-lives correctly
😢 Uh oh! Incorrect answer, Try again
Consider exponential decay
m = m₀(1/2)ⁿ
Calculate number of half-lives: 50/12.5 = 4
Use formula: m = m₀(1/2)ⁿ where n=4
Calculate: 64 × (1/16) = 4 mg
Not counting half-lives correctly
A particle is dropped from a height 'H'. The de'Broglie wavelength of the particle depends on height as
🥳 Wohoo! Correct answer
λ = h/mv
Use λ = h/mv and v = √(2gH)
Substitute v in wavelength formula
Get λ ∝ 1/√H or H⁻¹/²
Not considering velocity-height relationship
😢 Uh oh! Incorrect answer, Try again
Consider conservation of energy
λ = h/mv
Use λ = h/mv and v = √(2gH)
Substitute v in wavelength formula
Get λ ∝ 1/√H or H⁻¹/²
Not considering velocity-height relationship
The energy equivalent to a substance of mass 1 g is
🥳 Wohoo! Correct answer
E = mc²
Convert 1g to kg: m = 10⁻³ kg
Use Einstein's equation: E = mc²
Calculate: E = 10⁻³ × (3×10⁸)² = 9×10¹³ J
Not converting units
😢 Uh oh! Incorrect answer, Try again
Remember to convert units
E = mc²
Convert 1g to kg: m = 10⁻³ kg
Use Einstein's equation: E = mc²
Calculate: E = 10⁻³ × (3×10⁸)² = 9×10¹³ J
Not converting units
The period of revolution of an electron in the ground state of the hydrogen atom is T. The period of revolution of the electron in the first excited state is
🥳 Wohoo! Correct answer
T ∝ n³
Use T ∝ r³/² where r = n²a₀
For n=1 to n=2, ratio is (2²/1²)³/²
Calculate: T₂/T₁ = 8
Not cubing n
😢 Uh oh! Incorrect answer, Try again
Consider radius relationship with n
T ∝ n³
Use T ∝ r³/² where r = n²a₀
For n=1 to n=2, ratio is (2²/1²)³/²
Calculate: T₂/T₁ = 8
Not cubing n
The total energy of an electron revolving in the second orbit of the hydrogen atom is
🥳 Wohoo! Correct answer
E_n = -13.6/n² eV
Use Bohr's energy formula: E_n = -13.6/n² eV
For n=2, substitute in formula: E₂ = -13.6/2²
Calculate: E₂ = -13.6/4 = -3.4 eV
Not squaring n
😢 Uh oh! Incorrect answer, Try again
Remember energy is negative for bound states
E_n = -13.6/n² eV
Use Bohr's energy formula: E_n = -13.6/n² eV
For n=2, substitute in formula: E₂ = -13.6/2²
Calculate: E₂ = -13.6/4 = -3.4 eV
Not squaring n
The number of photons falling per second on a completely darkened plate to produce a force of 6.62×10⁻⁵ N is 'n'. If the wavelength of the light falling is 5×10⁻⁷ m, then n = _____×10²². (h = 6.62×10⁻³⁴ J-s)
🥳 Wohoo! Correct answer
F = (nhf)/c
Use radiation pressure formula: P = F/A = n(h/λ)/t
Rearrange to solve for n: n = (Fλ/h)t
Calculate: n = (6.62×10⁻⁵ × 5×10⁻⁷)/(6.62×10⁻³⁴) = 5×10²²
Forgetting momentum of photon
😢 Uh oh! Incorrect answer, Try again
Remember to include time factor
F = (nhf)/c
Use radiation pressure formula: P = F/A = n(h/λ)/t
Rearrange to solve for n: n = (Fλ/h)t
Calculate: n = (6.62×10⁻⁵ × 5×10⁻⁷)/(6.62×10⁻³⁴) = 5×10²²
Forgetting momentum of photon
The maximum kinetic energy of emitted photoelectrons depends on
🥳 Wohoo! Correct answer
K.E_max = hf - φ
Recall Einstein's photoelectric equation
Note that K.E_max = hf - φ
Observe dependence on frequency only
Confusing intensity with energy
😢 Uh oh! Incorrect answer, Try again
Consider energy conservation
K.E_max = hf - φ
Recall Einstein's photoelectric equation
Note that K.E_max = hf - φ
Observe dependence on frequency only
Confusing intensity with energy
Two protons are kept at a separation of 10nm. Let Fn and Fe be the nuclear force and the electromagnetic force between them
🥳 Wohoo! Correct answer
Fe ∝ 1/r², Fn ≈ 0 for r >> 10⁻¹⁵ m
Nuclear force range ≈ 10⁻¹⁵ m
At 10nm = 10⁻⁸ m, nuclear force negligible
Only Coulomb force significant
Not considering force ranges
😢 Uh oh! Incorrect answer, Try again
Compare force ranges
Fe ∝ 1/r², Fn ≈ 0 for r >> 10⁻¹⁵ m
Nuclear force range ≈ 10⁻¹⁵ m
At 10nm = 10⁻⁸ m, nuclear force negligible
Only Coulomb force significant
Not considering force ranges
A cyclotron's oscillator frequency is 10 MHz and the operating magnetic field is 0.66 T. If the radius of its dees is 60 cm, then the kinetic energy of the proton beam produced by the accelerator is
🥳 Wohoo! Correct answer
E = q²B²R²/2m
First use frequency equation: f = qB/2πm to verify field strength (10⁶ Hz = [1.6×10⁻¹⁹ C × 0.66 T]/[2π × 1.67×10⁻²⁷ kg])
Use energy equation: E = q²B²R²/2m where R = 0.6m
Calculate final energy in MeV: E = [1.6×10⁻¹⁹)² × (0.66)² × (0.6)²]/[2 × 1.67×10⁻²⁷] = 7 MeV
Not converting units
😢 Uh oh! Incorrect answer, Try again
Remember to convert units to MeV
E = q²B²R²/2m
First use frequency equation: f = qB/2πm to verify field strength (10⁶ Hz = [1.6×10⁻¹⁹ C × 0.66 T]/[2π × 1.67×10⁻²⁷ kg])
Use energy equation: E = q²B²R²/2m where R = 0.6m
Calculate final energy in MeV: E = [1.6×10⁻¹⁹)² × (0.66)² × (0.6)²]/[2 × 1.67×10⁻²⁷] = 7 MeV
Not converting units
The end product of decay of ⁹⁰Th²³² is ₈₂Pb²⁰⁸. The number of α and β particles emitted are respectively
🥳 Wohoo! Correct answer
ΔA = 4(#α), ΔZ = 2(#α) - (#β)
Change in mass number = 232 - 208 = 24
Each α particle reduces mass by 4 units
Need 6α and 4β to match atomic number
Not balancing both A and Z
😢 Uh oh! Incorrect answer, Try again
Balance nuclear equation
ΔA = 4(#α), ΔZ = 2(#α) - (#β)
Change in mass number = 232 - 208 = 24
Each α particle reduces mass by 4 units
Need 6α and 4β to match atomic number
Not balancing both A and Z
A hydrogen atom in ground state absorbs 10.2 eV of energy. The orbital angular momentum of the electron is increased by
🥳 Wohoo! Correct answer
L = nh/2π
Angular momentum in nth orbit = nh/2π
Change in L = Δ(nh/2π) = h/2π
Convert to Js: h/2π = 1.05x10⁻³⁴ Js
Unit conversion errors
😢 Uh oh! Incorrect answer, Try again
Use quantization of angular momentum
L = nh/2π
Angular momentum in nth orbit = nh/2π
Change in L = Δ(nh/2π) = h/2π
Convert to Js: h/2π = 1.05x10⁻³⁴ Js
Unit conversion errors
Frequency of revolution of an electron revolving in nth orbit of H-atom is proportional to
🥳 Wohoo! Correct answer
f = v/2πr
Use Bohr's model: v = √(ke²/mr)
r ∝ n², v ∝ 1/n
Frequency f = v/2πr ∝ 1/n³
Not relating radius to velocity
😢 Uh oh! Incorrect answer, Try again
Apply Bohr's atomic model
f = v/2πr
Use Bohr's model: v = √(ke²/mr)
r ∝ n², v ∝ 1/n
Frequency f = v/2πr ∝ 1/n³
Not relating radius to velocity
An electron is moving with an initial velocity V₀ = Vi ̂ and is in a uniform magnetic field B₀ = Bj ̂. Then its de Broglie wavelength
🥳 Wohoo! Correct answer
λ = h/mv, F = qv×B
λ = h/mv depends only on speed
Magnetic force is perpendicular to velocity
Speed remains constant, only direction changes
Thinking magnetic force changes speed
😢 Uh oh! Incorrect answer, Try again
Consider nature of magnetic force
λ = h/mv, F = qv×B
λ = h/mv depends only on speed
Magnetic force is perpendicular to velocity
Speed remains constant, only direction changes
Thinking magnetic force changes speed
In Rutherford experiment, for head-on collision of α particles with a gold nucleus, the impact parameter is
🥳 Wohoo! Correct answer
b = r sin(θ/2)
Impact parameter is perpendicular distance from initial path to target
For head-on collision, particle aims directly at nucleus
Therefore impact parameter = 0
Not understanding impact parameter
😢 Uh oh! Incorrect answer, Try again
Consider definition of impact parameter
b = r sin(θ/2)
Impact parameter is perpendicular distance from initial path to target
For head-on collision, particle aims directly at nucleus
Therefore impact parameter = 0
Not understanding impact parameter
Light of certain frequency and intensity incident on a photosensitive material causes photoelectric effect. If both the frequency and intensity are doubled, the photoelectric saturation current becomes
🥳 Wohoo! Correct answer
I_s ∝ intensity
Saturation current proportional to intensity
Frequency only affects electron energy
Double intensity means double current
Confusing frequency and intensity effects
😢 Uh oh! Incorrect answer, Try again
Remember intensity-current relationship
I_s ∝ intensity
Saturation current proportional to intensity
Frequency only affects electron energy
Double intensity means double current
Confusing frequency and intensity effects
During a β⁻ decay
🥳 Wohoo! Correct answer
n → p + e⁻ + ν̄ₑ
In β⁻ decay, a neutron converts to a proton by weak interaction: n → p + e⁻ + ν̄ₑ
The electron is not present in nucleus before decay but is created during decay process along with antineutrino.
This explains charge and mass number conservation in β⁻ decay.
Thinking electron exists in nucleus
😢 Uh oh! Incorrect answer, Try again
Remember weak interaction
n → p + e⁻ + ν̄ₑ
In β⁻ decay, a neutron converts to a proton by weak interaction: n → p + e⁻ + ν̄ₑ
The electron is not present in nucleus before decay but is created during decay process along with antineutrino.
This explains charge and mass number conservation in β⁻ decay.
Thinking electron exists in nucleus
A radio-active element has half-life of 15 years. What is the fraction that will decay in 30 years?
🥳 Wohoo! Correct answer
N = N₀(1/2)ⁿ
After one half-life (15 years), half of sample remains: N(15)/N₀ = 1/2. After two half-lives (30 years): N(30)/N₀ = (1/2)² = 1/4.
Fraction remaining = 1/4. Therefore, fraction decayed = 1 - 1/4 = 3/4 = 0.75.
This shows 75% of original sample has decayed in 30 years.
Not subtracting from 1
😢 Uh oh! Incorrect answer, Try again
Use exponential decay law
N = N₀(1/2)ⁿ
After one half-life (15 years), half of sample remains: N(15)/N₀ = 1/2. After two half-lives (30 years): N(30)/N₀ = (1/2)² = 1/4.
Fraction remaining = 1/4. Therefore, fraction decayed = 1 - 1/4 = 3/4 = 0.75.
This shows 75% of original sample has decayed in 30 years.
Not subtracting from 1
Two protons are kept at a separation of 10 nm. Let Fₙ and Fₑ the nuclear force and the electromagnetic force between them
🥳 Wohoo! Correct answer
Fₑ = kq²/r²
Nuclear force is strong but very short range (about 10⁻¹⁵ m). At 10 nm = 10⁻⁸ m, nuclear force is practically zero.
Coulomb force is long range and still significant at 10 nm. Calculate: Fₑ = kq²/r².
Therefore, Fₑ > Fₙ as electromagnetic force dominates at this large separation.
Not considering force ranges
😢 Uh oh! Incorrect answer, Try again
Consider range of nuclear forces
Fₑ = kq²/r²
Nuclear force is strong but very short range (about 10⁻¹⁵ m). At 10 nm = 10⁻⁸ m, nuclear force is practically zero.
Coulomb force is long range and still significant at 10 nm. Calculate: Fₑ = kq²/r².
Therefore, Fₑ > Fₙ as electromagnetic force dominates at this large separation.
Not considering force ranges
Angular momentum of an electron in hydrogen atom is 3h/2π (h is the Planck's constant). The K.E. of the electron is
🥳 Wohoo! Correct answer
KE = 13.6/n² eV
In Bohr model, angular momentum L = nh/2π where n is orbit number. Given L = 3h/2π, so n = 3.
For nth orbit, Total Energy = -13.6/n² eV and KE = +13.6/n² eV. For n = 3, calculate KE.
KE = 13.6/9 = 1.51 eV. Remember KE is positive and half the magnitude of total energy.
Confusing total energy and KE
😢 Uh oh! Incorrect answer, Try again
Remember relationship between n and energy
KE = 13.6/n² eV
In Bohr model, angular momentum L = nh/2π where n is orbit number. Given L = 3h/2π, so n = 3.
For nth orbit, Total Energy = -13.6/n² eV and KE = +13.6/n² eV. For n = 3, calculate KE.
KE = 13.6/9 = 1.51 eV. Remember KE is positive and half the magnitude of total energy.
Confusing total energy and KE
Three photodiodes D₁, D₂ and D₃ are made of semiconductors having band gaps of 2.5eV, 2eV and 3eV respectively. Which one will be able to detect light of wavelength 600 nm?
🥳 Wohoo! Correct answer
E = hc/λ, E > Eg for detection
First, calculate photon energy at 600nm: E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(600×10⁻⁹) = 2.07 eV
Compare this energy with band gaps: For detection, photon energy must exceed band gap. 2.07 eV is greater than D₂'s band gap (2eV) but less than D₁ (2.5eV) and D₃ (3eV).
Therefore, only D₂ can detect this light as it has sufficient energy to excite electrons across its band gap.
Forgetting energy threshold
😢 Uh oh! Incorrect answer, Try again
Photon energy must exceed band gap
E = hc/λ, E > Eg for detection
First, calculate photon energy at 600nm: E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(600×10⁻⁹) = 2.07 eV
Compare this energy with band gaps: For detection, photon energy must exceed band gap. 2.07 eV is greater than D₂'s band gap (2eV) but less than D₁ (2.5eV) and D₃ (3eV).
Therefore, only D₂ can detect this light as it has sufficient energy to excite electrons across its band gap.
Forgetting energy threshold
The period of revolution of an electron revolving in nth orbit of H-atom is proportional to
🥳 Wohoo! Correct answer
T = 2πr/v
Use Bohr's model: r ∝ n², v ∝ 1/n. Period T = 2πr/v.
Substitute the relationships: T = 2π(kn²)/(k'/n) where k, k' are constants.
Simplify to get T ∝ n³. Period increases as cube of principal quantum number.
Not using correct Bohr model relations
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Remember Bohr's relationships
T = 2πr/v
Use Bohr's model: r ∝ n², v ∝ 1/n. Period T = 2πr/v.
Substitute the relationships: T = 2π(kn²)/(k'/n) where k, k' are constants.
Simplify to get T ∝ n³. Period increases as cube of principal quantum number.
Not using correct Bohr model relations
The de-Broglie wavelength associated with electron of hydrogen atom in this ground state is
🥳 Wohoo! Correct answer
λ = h/mv
For ground state of hydrogen, E = -13.6 eV. Use de Broglie wavelength formula: λ = h/p = h/mv.
Find velocity from energy: ½mv² = 13.6 eV. Convert to appropriate units and solve for v.
Calculate wavelength using λ = h/mv. Result is 3.3 Å.
Unit conversion errors
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Remember energy in ground state
λ = h/mv
For ground state of hydrogen, E = -13.6 eV. Use de Broglie wavelength formula: λ = h/p = h/mv.
Find velocity from energy: ½mv² = 13.6 eV. Convert to appropriate units and solve for v.
Calculate wavelength using λ = h/mv. Result is 3.3 Å.
Unit conversion errors
A hot filament liberates an electron with zero initial velocity. The anode potential is 1200V. The speed of electron when it strikes anode is
🥳 Wohoo! Correct answer
eV = ½mv²
First understand that the electron starts with zero velocity and gains kinetic energy from the electric field. The work done by electric field (eV) equals the final kinetic energy (½mv²).
Carefully substitute values: Electric potential energy = eV = (1.6×10⁻¹⁹ C)(1200 V), mass of electron = 9.1×10⁻³¹ kg. Set this equal to ½mv².
Solve for velocity: v = √(2eV/m) = √(2×1.6×10⁻¹⁹×1200)/(9.1×10⁻³¹) = 2.1×10⁷ m/s
Forgetting to use electron mass and charge values
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Energy gained from electric field converts to kinetic energy
eV = ½mv²
First understand that the electron starts with zero velocity and gains kinetic energy from the electric field. The work done by electric field (eV) equals the final kinetic energy (½mv²).
Carefully substitute values: Electric potential energy = eV = (1.6×10⁻¹⁹ C)(1200 V), mass of electron = 9.1×10⁻³¹ kg. Set this equal to ½mv².
Solve for velocity: v = √(2eV/m) = √(2×1.6×10⁻¹⁹×1200)/(9.1×10⁻³¹) = 2.1×10⁷ m/s
Forgetting to use electron mass and charge values
In a nuclear reactor heavy nuclei is not used as moderators because
🥳 Wohoo! Correct answer
ΔE ∝ 4M₁M₂/(M₁+M₂)²
Let's understand the purpose of moderators first: they slow down fast neutrons to thermal energies for sustaining chain reaction.
In elastic collisions, maximum energy transfer occurs when masses are similar. For neutrons hitting heavy nuclei, very little energy is transferred per collision (like a ping pong ball hitting a bowling ball).
Light nuclei like hydrogen (in water) or carbon (in graphite) are much more effective moderators because their mass is closer to neutron mass, allowing more energy transfer per collision.
Thinking heavier nuclei are better moderators
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Maximum energy transfer occurs when colliding masses are similar
ΔE ∝ 4M₁M₂/(M₁+M₂)²
Let's understand the purpose of moderators first: they slow down fast neutrons to thermal energies for sustaining chain reaction.
In elastic collisions, maximum energy transfer occurs when masses are similar. For neutrons hitting heavy nuclei, very little energy is transferred per collision (like a ping pong ball hitting a bowling ball).
Light nuclei like hydrogen (in water) or carbon (in graphite) are much more effective moderators because their mass is closer to neutron mass, allowing more energy transfer per collision.
Thinking heavier nuclei are better moderators
An electron in an excited state of Li²⁺ ion has angular momentum 3h/2π. The de Broglie wavelength... [continue]
🥳 Wohoo! Correct answer
L = mvr = nh/2π
Use angular momentum quantization: mvr = 3h/2π. For Li²⁺, Z=3.
De Broglie wavelength λ = h/mv = h/(3h/2πr) = 2πr/3
Given r = a₀/Z where a₀ is Bohr radius. Substitute and solve to get P = 2.
Confusion in Z dependence
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Angular momentum determines orbit radius
L = mvr = nh/2π
Use angular momentum quantization: mvr = 3h/2π. For Li²⁺, Z=3.
De Broglie wavelength λ = h/mv = h/(3h/2πr) = 2πr/3
Given r = a₀/Z where a₀ is Bohr radius. Substitute and solve to get P = 2.
Confusion in Z dependence
Energy of an electron in the second orbit of hydrogen atom is E₂. The energy of electron in the third orbit of He⁺ will be
🥳 Wohoo! Correct answer
En = -13.6(Z²/n²) eV
For hydrogen-like atoms, En = -13.6(Z²/n²) eV. For hydrogen in n=2, E₂ = -13.6/4 eV.
For He⁺, Z=2 and n=3. E = -13.6(4/9) eV
Taking ratio of He⁺(n=3) to H(n=2): (-13.6×4/9)/(-13.6/4) = 16/9
Forgetting Z dependence
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Energy depends on Z² and 1/n²
En = -13.6(Z²/n²) eV
For hydrogen-like atoms, En = -13.6(Z²/n²) eV. For hydrogen in n=2, E₂ = -13.6/4 eV.
For He⁺, Z=2 and n=3. E = -13.6(4/9) eV
Taking ratio of He⁺(n=3) to H(n=2): (-13.6×4/9)/(-13.6/4) = 16/9
Forgetting Z dependence
The work function of a metal is 1eV. Light of wavelength 3000Å is incident on this metal surface. The velocity of emitted photoelectrons will be
🥳 Wohoo! Correct answer
KE = hν - φ, ½mv² = KE
First, calculate photon energy: E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(3000×10⁻¹⁰) = 4.14 eV
Using Einstein's photoelectric equation: KE = hν - φ = 4.14 - 1 = 3.14 eV. Convert to Joules: 3.14×1.6×10⁻¹⁹ J
For velocity: ½mv² = KE. Solving for v = √(2KE/m) ≈ 10⁶ m/s
Unit conversion errors
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Energy conservation between photon and electron
KE = hν - φ, ½mv² = KE
First, calculate photon energy: E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(3000×10⁻¹⁰) = 4.14 eV
Using Einstein's photoelectric equation: KE = hν - φ = 4.14 - 1 = 3.14 eV. Convert to Joules: 3.14×1.6×10⁻¹⁹ J
For velocity: ½mv² = KE. Solving for v = √(2KE/m) ≈ 10⁶ m/s
Unit conversion errors
Which radiation deflected by electric field?:
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F = qE
Consider particle nature and charge of each: γ-rays (EM waves), α-particles (He²⁺), X-rays (EM waves), neutrons (neutral)
Only charged particles are deflected by electric field
α-particles are positively charged, so they're deflected! Others aren't affected
Thinking all radiation deflects
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Consider particle charge
F = qE
Consider particle nature and charge of each: γ-rays (EM waves), α-particles (He²⁺), X-rays (EM waves), neutrons (neutral)
Only charged particles are deflected by electric field
α-particles are positively charged, so they're deflected! Others aren't affected
Thinking all radiation deflects
Nuclear reactor 9×10⁹W power, fuel consumed per hour is:
🥳 Wohoo! Correct answer
P = mc²/t
Use E = mc². Power is energy per time, so P = (mc²)/t
Given P = 9×10⁹W = 9×10⁹J/s, t = 3600s (1 hour)
Solve for m: m = (Pt)/(c²) = 0.04g. Small mass produces huge energy!
Unit conversion errors
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Remember mass-energy equivalence
P = mc²/t
Use E = mc². Power is energy per time, so P = (mc²)/t
Given P = 9×10⁹W = 9×10⁹J/s, t = 3600s (1 hour)
Solve for m: m = (Pt)/(c²) = 0.04g. Small mass produces huge energy!
Unit conversion errors
Photoelectron KE increases 0.52eV when wavelength changes from 500nm to new wavelength. New wavelength is:
🥳 Wohoo! Correct answer
ΔKE = hc(1/λ₂ - 1/λ₁)
Use Einstein's photoelectric equation: KE = hc(1/λ₂ - 1/λ₁)
Given ΔKE = 0.52eV = hc(1/λ₂ - 1/500nm)
Solve for λ₂, getting 400nm. Shorter wavelength means higher energy!
Wavelength-energy confusion
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Remember energy-wavelength relation
ΔKE = hc(1/λ₂ - 1/λ₁)
Use Einstein's photoelectric equation: KE = hc(1/λ₂ - 1/λ₁)
Given ΔKE = 0.52eV = hc(1/λ₂ - 1/500nm)
Solve for λ₂, getting 400nm. Shorter wavelength means higher energy!
Wavelength-energy confusion
Electron in Bohr orbit radius 0.529Å, radius of third orbit is:
🥳 Wohoo! Correct answer
r_n = n²r₁
In Bohr model, radius of nth orbit is r_n = n²r₁, where r₁ is ground state radius
For n=3: r₃ = (3²)(0.529Å) = (9)(0.529Å)
Calculate: r₃ = 4.761Å. Keep units consistent!
Unit conversion errors
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Use n² dependence
r_n = n²r₁
In Bohr model, radius of nth orbit is r_n = n²r₁, where r₁ is ground state radius
For n=3: r₃ = (3²)(0.529Å) = (9)(0.529Å)
Calculate: r₃ = 4.761Å. Keep units consistent!
Unit conversion errors
In photoelectric effect, if intensity and frequency doubled, saturation current:
🥳 Wohoo! Correct answer
i_s ∝ intensity
Saturation current depends on number of electrons emitted per second
Intensity determines number of photoelectrons (doubled), frequency affects KE
When both double, only intensity affects current, so current doubles!
Confusing intensity and frequency effects
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Think about intensity's role
i_s ∝ intensity
Saturation current depends on number of electrons emitted per second
Intensity determines number of photoelectrons (doubled), frequency affects KE
When both double, only intensity affects current, so current doubles!
Confusing intensity and frequency effects
Binding energy of ¹⁴N, mass=14.00307u is:
🥳 Wohoo! Correct answer
BE = Δm×931.5 MeV/u
Binding energy = [Zm_p + (A-Z)m_n - M]c² where M is actual mass
For ¹⁴N: Z=7, A=14. Calculate mass defect using given mass
Convert to energy using E=mc². BE = 104.7 MeV. Check units!
Mass unit conversion
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Consider mass defect
BE = Δm×931.5 MeV/u
Binding energy = [Zm_p + (A-Z)m_n - M]c² where M is actual mass
For ¹⁴N: Z=7, A=14. Calculate mass defect using given mass
Convert to energy using E=mc². BE = 104.7 MeV. Check units!
Mass unit conversion
De Broglie wavelength λ at kinetic energy K, wavelength at K/4 is:
🥳 Wohoo! Correct answer
λ = h/√(2mK)
de Broglie wavelength λ = h/p = h/√(2mK). Let's write this for both energies
For K/4: λ₂ = h/√(2m(K/4)) = h/√(mK/2)
Compare with original λ₁ = h/√(2mK): λ₂/λ₁ = √2 = 2. So λ₂ = 2λ₁!
Forgetting square root
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Remember λ ∝ 1/√K
λ = h/√(2mK)
de Broglie wavelength λ = h/p = h/√(2mK). Let's write this for both energies
For K/4: λ₂ = h/√(2m(K/4)) = h/√(mK/2)
Compare with original λ₁ = h/√(2mK): λ₂/λ₁ = √2 = 2. So λ₂ = 2λ₁!
Forgetting square root
H-atom ground state radius 0.53Å, after collision radius 2.12Å, final state n is:
🥳 Wohoo! Correct answer
r = n²a₀
In Bohr model, radius ∝ n². Let's set up ratio: r₂/r₁ = (n₂/n₁)²
Given r₁ = 0.53Å (n₁=1), r₂ = 2.12Å. Substitute: 2.12/0.53 = n₂²
Solve: 4 = n₂², therefore n₂ = 2. The electron jumped to second orbit!
Forgetting square relationship
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Use ratio method
r = n²a₀
In Bohr model, radius ∝ n². Let's set up ratio: r₂/r₁ = (n₂/n₁)²
Given r₁ = 0.53Å (n₁=1), r₂ = 2.12Å. Substitute: 2.12/0.53 = n₂²
Solve: 4 = n₂², therefore n₂ = 2. The electron jumped to second orbit!
Forgetting square relationship
A radioactive sample has half-life of 3 years. The time required for the activity of the sample to reduce to 1/5th of its initial value is about
🥳 Wohoo! Correct answer
N=N₀e⁻λt
Use N=N₀e⁻λt
For 1/5th: 1/5=e⁻λt
t=(ln5)/λ≈7 years
Half-life relation
😢 Uh oh! Incorrect answer, Try again
Consider decay constant
N=N₀e⁻λt
Use N=N₀e⁻λt
For 1/5th: 1/5=e⁻λt
t=(ln5)/λ≈7 years
Half-life relation
The energy gap of an LED is 2.4eV. When the LED is switched 'ON', the momentum of the emitted photons is
🥳 Wohoo! Correct answer
p=E/c
Convert energy to joules: E=2.4×1.6×10⁻¹⁹J
Use p=E/c for photon momentum
p=1.28×10⁻²⁷ kg.m.s⁻¹
Unit conversion
😢 Uh oh! Incorrect answer, Try again
Consider photon momentum
p=E/c
Convert energy to joules: E=2.4×1.6×10⁻¹⁹J
Use p=E/c for photon momentum
p=1.28×10⁻²⁷ kg.m.s⁻¹
Unit conversion
A nucleus with mass number 220 initially at rest emits an alpha particle. If the Q value of reaction is 5.5MeV, calculate the value of kinetic energy of alpha particle
🥳 Wohoo! Correct answer
KE=(A-4/A)Q
Use momentum conservation
KE=(A-4/A)Q
KE=(216/220)×5.5=5.4MeV
Energy distribution
😢 Uh oh! Incorrect answer, Try again
Consider mass ratios
KE=(A-4/A)Q
Use momentum conservation
KE=(A-4/A)Q
KE=(216/220)×5.5=5.4MeV
Energy distribution
In the following equation representing β⁻ decay, the number of neutrons in the nucleus X is ⁸³²¹⁰Bi→X+e⁻+v̄
🥳 Wohoo! Correct answer
n=A-Z
Write nuclear equation
Calculate neutrons using mass number
n=A-Z=210-84=126
Mass number confusion
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Consider beta decay
n=A-Z
Write nuclear equation
Calculate neutrons using mass number
n=A-Z=210-84=126
Mass number confusion
A 60 W source emits monochromatic light of wavelength 662.5 nm. The number of photons emitted per second is
🥳 Wohoo! Correct answer
E=hc/λ
Calculate energy per photon: E=hc/λ
Find number using P=nE/t
n=Pt/E=2×10²⁰ photons/s
Unit conversion
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Consider energy conservation
E=hc/λ
Calculate energy per photon: E=hc/λ
Find number using P=nE/t
n=Pt/E=2×10²⁰ photons/s
Unit conversion
The photoelectric work function for photo metal is 2.4eV...
🥳 Wohoo! Correct answer
hc/λ=φ+KEmax
Use Einstein's equation: hc/λ=φ
Calculate threshold wavelength: λ₀=12400/2.4=5166Å
700nm>516.6nm, no emission
Wavelength threshold
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Check threshold
hc/λ=φ+KEmax
Use Einstein's equation: hc/λ=φ
Calculate threshold wavelength: λ₀=12400/2.4=5166Å
700nm>516.6nm, no emission
Wavelength threshold
In the Rutherford's alpha scattering experiment, as the impact parameter increases, the scattering angle of the alpha particle
🥳 Wohoo! Correct answer
θ ∝ 1/b
Large impact parameter means particle passes farther
Force decreases with distance
Scattering angle decreases
Distance effect confusion
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Consider force variation
θ ∝ 1/b
Large impact parameter means particle passes farther
Force decreases with distance
Scattering angle decreases
Distance effect confusion
Light of energy E falls normally on metal of work function E/3...
🥳 Wohoo! Correct answer
E=hν=φ+KEmax
Apply photoelectric equation: E=φ+KEmax
Substitute φ=E/3 to get KEmax=2E/3
KE range is 0 to 2E/3
Energy distribution
😢 Uh oh! Incorrect answer, Try again
Consider energy range
E=hν=φ+KEmax
Apply photoelectric equation: E=φ+KEmax
Substitute φ=E/3 to get KEmax=2E/3
KE range is 0 to 2E/3
Energy distribution
Consider nuclear fission reaction ¹n+²³⁵U→¹⁴⁴Ba+⁸⁹Kr+3¹n...
🥳 Wohoo! Correct answer
ΔE=ΔBE
Calculate total binding energy difference: (1980-1800)MeV
Energy per neutron=(180)/3 MeV
Each neutron carries 60 MeV
Energy distribution
😢 Uh oh! Incorrect answer, Try again
Consider energy conservation
ΔE=ΔBE
Calculate total binding energy difference: (1980-1800)MeV
Energy per neutron=(180)/3 MeV
Each neutron carries 60 MeV
Energy distribution
The ratio of area of first excited state to ground state of orbit of hydrogen atom is
🥳 Wohoo! Correct answer
A∝πr²∝n⁴
Use Bohr's theory: radius of orbit r∝n² where n is principal quantum number
Area of orbit ∝r² ∝n⁴
Compare n=2 (first excited) to n=1 (ground): (2⁴:1⁴)=16:1
Radius vs area relation
😢 Uh oh! Incorrect answer, Try again
Consider quantum numbers
A∝πr²∝n⁴
Use Bohr's theory: radius of orbit r∝n² where n is principal quantum number
Area of orbit ∝r² ∝n⁴
Compare n=2 (first excited) to n=1 (ground): (2⁴:1⁴)=16:1
Radius vs area relation
The ratio of volume of Al²⁷ nucleus to its surface area is (Given R₀=1.2×10⁻¹⁵m)
🥳 Wohoo! Correct answer
R=R₀A¹/³
Use R=R₀A¹/³ for nuclear radius
Calculate ratio: V/A=(4πR³/3)/(4πR²)=R/3
Substitute R=R₀(27)¹/³=1.2×10⁻¹⁵m
Volume/area relation
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Consider nuclear size
R=R₀A¹/³
Use R=R₀A¹/³ for nuclear radius
Calculate ratio: V/A=(4πR³/3)/(4πR²)=R/3
Substitute R=R₀(27)¹/³=1.2×10⁻¹⁵m
Volume/area relation
In alpha particle scattering experiment, if v is initial velocity...
🥳 Wohoo! Correct answer
d∝1/v²
Use relationship d∝1/v² from Rutherford formula
When v→2v, d→d/4
New distance is d/4
Velocity relation
😢 Uh oh! Incorrect answer, Try again
Consider inverse square
d∝1/v²
Use relationship d∝1/v² from Rutherford formula
When v→2v, d→d/4
New distance is d/4
Velocity relation