JEE Physics Modern Physics MCQs Set 05

Answer: (d) 10.24 MeV

Explanation:
1. Four He nuclei form one \( O-16 \) nucleus: \( \Delta m = 4(4.0026)-15.9994 = 0.0110 u \).
2. \( E = 0.0110\times931.5 = 10.24 \) MeV.

Teacher's Note:
1. Four helium nuclei make one oxygen nucleus.
2. \( 1 u = 931.5 \) MeV.

Answer: (d) 7.1 MeV

Explanation:
1. Binding energy \( = 0.0305\times931 = 28.4 \) MeV.
2. Per nucleon \( = \frac{28.4}{4} = 7.1 \) MeV.

Teacher's Note:
1. Divide the total by \( A = 4 \).
2. \( 1 u = 931 \) MeV.

Answer: (a) \( M = Nm_n+Zm_p-\frac{B}{c^2} \)

Explanation:
1. The nucleus is lighter than its separated nucleons by the mass equivalent of the binding energy.
2. Mass defect \( = \frac{B}{c^2} \).

Teacher's Note:
1. Check dimensions: \( B/c^{2} \) is a mass.
2. Mass defect is subtracted.

Answer: (c) \( c-a-b \)

Explanation:
1. Energy released = binding energy of products − binding energy of reactants.
2. The neutron has no binding energy, so \( Q = c - (a + b) \).

Teacher's Note:
1. More binding means more energy released.
2. Free nucleons have zero binding energy.

Answer: (a) 5.4 MeV

Explanation:
1. Momentum conservation gives \( K_\alpha = Q\frac{A-4}{A} \).
2. \( K_\alpha = 5.5\times\frac{216}{220} = 5.4 \) MeV.

Teacher's Note:
1. Lighter product gets the larger share of \( K.E \).
2. Shares are inversely proportional to mass.

Answer: (d) \( \frac{A-Z-4}{Z-8} \)

Explanation:
1. Final mass number = \( A - 12 \). Protons: \( Z - 6 (\alpha) - 2 \) (positron decay) = \( Z - 8 \).
2. Neutrons = \( (A - 12) - (Z - 8) = A - Z - 4 \).

Teacher's Note:
1. A positron emission lowers \( Z \) by 1.
2. Neutrons = mass number − protons.

Answer: (b) 40 min

Explanation:
1. \( t = T\log_2\frac{N_0}{N} \): at \( t_{2} \) the fraction left is 1/5, so \( t_2 = 20\log_2 5 \); at \( t_{1} \) the fraction left is 4/5, so \( t_1 = 20\log_2\frac54 \).
2. \( t_2-t_1 = 20\log_2 4 = 40 \min \).

Teacher's Note:
1. Remaining fractions are 1/5 and 4/5.
2. Their ratio is 4, two half-lives.

Answer: (c) \( N_1\lambda_1e^{-\lambda_1t}+N_2\lambda_2e^{-\lambda_2t} \)

Explanation:
1. Each species decays independently at rate \( \lambda N_0e^{-\lambda t} \).
2. The total rate is the sum.

Teacher's Note:
1. Rates of independent samples add.
2. Activity = \( \lambdaN \).

Answer: (a) zero

Explanation:
1. The end product of a series is stable and does not decay.
2. Hence its decay constant is zero.

Teacher's Note:
1. Stable nuclei do not disintegrate.
2. \( \lambda = 0 \) means infinite half-life.

Answer: (b) 125 years

Explanation:
1. Fraction of \( X \) left = \( 1/32 = (1/2)^{5} \).
2. Five half-lives: \( 5 \times 25 = 125 \) years.

Teacher's Note:
1. Remaining fraction is \( X/(X + Y) \).
2. Count the half-lives.

Answer: (d) electrons produced as a result of the decay of neutrons inside the nucleus

Explanation:
1. In β⁻ decay a neutron converts to a proton: \( n\to p+e^-+\bar\nu \).
2. Electrons are created at the time of decay and do not pre-exist in the nucleus.

Teacher's Note:
1. The nucleus has only protons and neutrons.
2. The emitted electron is created in the process.

Answer: (c) 2 microgram

Explanation:
1. Energy per minute \( = 3\times10^6\times60 = 1.8\times10^8 J \).
2. \( \Delta m = \frac{E}{c^2} = \frac{1.8\times10^8}{9\times10^{16}} = 2\times10^{-9} \) kg = \( 2 \mug \).

Teacher's Note:
1. Use \( E \) = mc².
2. Convert kW to \( W \) and minutes to seconds.

Answer: (c) 0.37 g

Explanation:
1. After one mean life the amount is \( \frac1e \) of the original.
2. After two: \( 2.718e^{-2} = 0.368 \approx 0.37 g \).

Teacher's Note:
1. Mean life τ = 1/λ.
2. \( N = N_{0}e^{-t/\tau } \).

Answer: (a) \( 1.75\times10^{15} \)

Explanation:
1. 60 days is 3 half-lives, so 1/8 of the nuclei remain.
2. Decayed = \( 2\times10^{15}\times\frac78 = 1.75\times10^{15} \).

Teacher's Note:
1. Decayed = initial − remaining.
2. 3 half-lives leave 1/8.

Answer: (b) \( {}^A_ZX^*\to{}^A_ZX+\gamma \)

Explanation:
1. In γ-decay an excited nucleus drops to a lower energy state emitting a photon, with \( Z \) and \( A \) unchanged.

Teacher's Note:
1. Only energy changes in γ decay.
2. Other options change \( Z \) or \( A \).

Answer: (b) 20 years and 5 years respectively

Explanation:
1. \( \lambda_1N_1 = 5 \), \( \lambda_2N_2 = 10 \), \( N_1 = 2N_2 \Rightarrow \frac{\lambda_1}{\lambda_2} = \frac14 \).
2. So \( T_{1} = 4T_{2} \): 20 years and 5 years.

Teacher's Note:
1. Activity = \( \lambdaN \).
2. Half-life is inversely proportional to λ.

Answer: (a) \( \lambda = \frac{h}{m\sqrt{2as}} \)

Explanation:
1. \( v^2 = 2as \Rightarrow v = \sqrt{2as} \).
2. \( \lambda = \frac{h}{mv} = \frac{h}{m\sqrt{2as}} \).

Teacher's Note:
1. Use \( v^{2} \) = 2as.
2. λ = h/mv.

Answer: (d) \( 1.5\times10^{-2} \)

Explanation:
1. \( \lambda_{ph} = 3\lambda_e = \frac{3h}{mv} \), so \( E_{ph} = \frac{hc}{\lambda_{ph}} = \frac{mvc}{3} \).
2. \( E_e = \tfrac12mv^2 \), so \( \frac{E_e}{E_{ph}} = \frac{3v}{2c} = 1.5\times10^{-2} \).

Teacher's Note:
1. Express both energies through mv.
2. \( v/c = 0.01 \).

Answer: (b) \( {}^{30}_{15}\mathrm{P} \)

Explanation:
1. Mass number: \( 4 + 27 = A + 1 \) so \( A = 30 \). Charge: \( 2 + 13 = Z + 0 \) so \( Z = 15 \).
2. \( X \) is phosphorus-30.

Teacher's Note:
1. Conserve \( A \) and \( Z \).
2. The neutron has \( Z = 0 \).

Answer: (a) 1 : 1

Explanation:
1. Momentum conservation gives equal and opposite momenta.
2. \( \lambda = \frac hp \) is therefore the same for both.

Teacher's Note:
1. Initial momentum is zero.
2. λ depends only on \( |p| \).

Answer: (c) 24

Explanation:
1. \( \Delta m = 2(2.0141)-4.0024 = 0.0258 u \).
2. \( Q = 0.0258\times931.5 \approx 24 \) MeV.

Teacher's Note:
1. Use the mass defect.
2. \( 1 u = 931.5 \) MeV.

Answer: (a) \( 1.5\times10^7 \) W

Explanation:
1. Force on a perfect reflector \( F = \frac{2P}{c} = mg = 0.1 N \).
2. \( P = \frac{0.1\times3\times10^8}{2} = 1.5\times10^7 W \).

Teacher's Note:
1. Reflection doubles the momentum change.
2. Balance with weight.

Answer: (d) \( 9.3\times10^{-8} \) N

Explanation:
1. \( F = \frac Pc(0.6\times1+0.4\times2) = \frac{20}{3\times10^8}\times1.4 \).
2. \( F = 9.3\times10^{-8} N \).

Teacher's Note:
1. Absorbed light transfers \( P/c \), reflected \( 2P/c \).
2. Weight the fractions.

Answer: (b) 3.3 m/s

Explanation:
1. Momentum conservation: \( mv = \frac h\lambda \).
2. \( v = \frac{6.63\times10^{-34}}{122\times10^{-9}\times1.6\times10^{-27}} = 3.4 m/s \), i.e. about \( 3.3 m/s \).

Teacher's Note:
1. Photon momentum is \( h/\lambda \).
2. The atom's momentum is cancelled.

Answer: (c) \( 9\lambda \)

Explanation:
1. \( \frac{hc}{\lambda}-\phi = 4eV_0 \) and \( \frac{hc}{3\lambda}-\phi = eV_0 \).
2. Subtracting: \( \frac{2hc}{3\lambda} = 3eV_0 \Rightarrow \frac{hc}{\lambda} = 4.5eV_0 \), so \( \phi = 0.5eV_0 \) and \( \lambda_0 = \frac{hc}{\phi} = 9\lambda \).

Teacher's Note:
1. Write Einstein's equation twice and subtract.
2. λ₀ = hc/φ.