IBDP Physics 12.2 – Nuclear physics: IB Style Question Bank HL Paper 1

Question

What was a reason to postulate the existence of neutrinos?

    1. Nuclear energy levels had a continuous spectrum.

    2. The photon emission spectrum only contained specific wavelengths.

    3. Some particles were indistinguishable from their antiparticle.

    4. The energy of emitted beta particles had a continuous spectrum.

Answer/Explanation

Ans: D

Wolfgang Pauli hypothesised the existence of a third particle in the products of a beta decay in 1933. Since the energy of the electron in beta decay has a range of possible values, it means that a third very light particle must also be produced so that it carries the remainder of the available energy. Enrico Fermi coined the word neutrino for the ‘little neutral one.

Question

The graphs show the variation with time of the activity and the number of remaining nuclei for a sample of a radioactive nuclide.

                               

What is the decay constant of the nuclide?

A. 0.7 {s}^{-1}

B. 1 {s}^{-1}

C. \tfrac{1}{0.7} {s}^{-1}

D. 1.5 {s}^{-1}

Answer/Explanation

Ans: A

Activity : The number of decays per unit time or decay rate is called activity(R)
where
From given graph at t= 0 
\(N_0=1\times 10^6, Activity = 0.7 \times 10^6\)
Hence putting these values in above equation , we get
\(\left [ R \right ]_0=N_0 \lambda e^{-\lambda t}\)
\(=N_0 \lambda e^{-\lambda \times 0} =N_0 \lambda\)
Hence
\(\lambda =\frac{\left [ R \right ]_0}{N_0}\)
\(=\frac{0.7\times 10^6}{1\times 10^6}\)
\(=0.7\)

Question

Two samples X and Y of different radioactive isotopes have the same initial activity. Sample X has twice the number of atoms as sample Y. The half-life of X is T. What is the half-life of Y?

A. 2T

B. T

C. \(\frac{T}{2}\)

D. \(\frac{T}{4}\)

Answer/Explanation

Markscheme

C

Initial activity

\(\left [ R \right ]_0=N_0 \lambda e^{-\lambda t}\)
\(=N_0 \lambda e^{-\lambda \times 0} =N_0 \lambda\)
Hence
\(\lambda =\frac{\left [ R \right ]_0}{N_0}\)

Sample X and Y has same initial activity Also 

\(N_{OX}=2N_{OY}\)
and
\(T_{1/2(X)}= T T_{1/2(Y)}= ?\)
\(N_{OX}\lambda_X=N_{OY}\lambda_Y\)
\(\lambda =\frac{0.63}{T_{1/2}}\)
hence
\(\frac{N_{OX}}{T_{1/2(X)}} =\frac{N_{OY}}{T_{1/2(Y)}}\)
or
\(T_{1/2(Y)} =\frac{N_{OY}}{N_{OX}}\times T_{1/2(X)}\)
\(=\frac{T_{1/2(X)}}{2}=\frac{T}{2}\)

Question

An electron of initial energy E tunnels through a potential barrier. What is the energy of the electron after tunnelling?

A. greater than E

B. E

C. less than E

D. zero

Answer/Explanation

Markscheme

B

Energy of electron remain same.

Question

Two radioactive nuclides, X and Y, have half-lives of 50 s and 100 s respectively. At time t = 0 samples of X and Y contain the same number of nuclei.

What is \(\frac{{{\text{number of nuclei of X undecayed}}}}{{{\text{number of nuclei of Y undecayed}}}}\) when t = 200 s?

A. 4

B. 2

C. \(\frac{1}{2}\)

D. \(\frac{1}{4}\)

Answer/Explanation

Markscheme

D

\(N_t=\frac{N_o}{2^n}\) where n is number of half life
For nuclides X  \(n_x=\frac{200}{50}=4\)
for nuclides Y \(n_y=\frac{200}{100}=2\)
Also given that  \(N_{0X} =N_{0X}\)
Hence
\(\frac{ N_{tX}}{ N_{tY}}=\frac{\frac{N_o}{2^{n_x}}}{\frac{N_o}{2^{n_y}}}\)
\(=\frac{2^2}{2^4}=\frac{1}{4}\)

Question

Alpha particles with energy E are directed at nuclei with atomic number Z. Small deviations from the predictions of the Rutherford scattering model are observed.

Which change in E and which change in Z is most likely to result in greater deviations from the Rutherford scattering model?

Answer/Explanation

Markscheme

B

The derivation of the Rutherford formula is based on a number of assumptions. The most important is that the only force in play during the scattering process is the electric force. As the energy of the alpha particles increases, the alpha particles can get closer to the nucleus. When the distance of closest approach gets to be about 10−15 m or less, deviations from the Rutherford formula are observed . This is due to the fact that the alpha particles are so close to the nucleus that the strong nuclear force begins to act on the alpha particles. Hence nuclear force is felt with high energy and smaller  Z .

Question

A particle of fixed energy is close to a potential barrier.

Which changes to the width of the barrier and to the height of the barrier will always make the tunnelling probability greater?

Answer/Explanation

Markscheme

D

An energy diagram containing two plots for the situation of Fig. above:

  1. The electron’s mechanical energy E is plotted when the electron is at any coordinate x < 0.
  2. The electron’s electric potential energy U is plotted as a function of the electron’s position x, assuming that the electron can reach any value of x.
  3. The nonzero part of the plot (the potential barrier) has height Ub and thickness L.

For a particle with mass m and a barrier of thickness L, the transmission coefficient is

\(T\approx e^{-2bL}\)

where
\(b=\sqrt{\frac{8\pi ^2 m(U_b-E)}{h^2}}\)

hence for Smaller \(L\) and \(U_b-E\) , \(T\) is greater

Question

A radioactive element has decay constant \(\lambda \) (expressed in s–1). The number of nuclei of this element at t = 0 is N. What is the expected number of nuclei that will have decayed after 1 s?

A.  \(N\left( {1 – {e^{ – \lambda }}} \right)\)

B.  \(\frac{N}{\lambda }\)

C.  \(N{e^{ – \lambda }}\)

D.  \(\lambda N\)

Answer/Explanation

Markscheme

A

Number of nuclei after time \(t\)  is given by
\(N_t= N_0 e^{-\lambda t}\) 
Now after 1 sec
\(N_1=N e^{-\lambda}\) 
hence Number of Nuclei declayed
\(= N – N_1\) 
\(=N-N e^{-\lambda}\) 
\(=N(1-e^{-\lambda})\)