Problem set 68

1. In a counting experiment to determine the statistics obeyed by the $\beta$ particles emitted by a radioactive substance, the number of $\beta$ particles counted in 50 seconds time interval was repeatedly measured 100 times. The statistical ensemble in this case consists of the following number of members:
1. 50
2. 100
3. 5000
4. 2

As the reading is repeated 100 times, there will be 100 members in the ensembles.

Hence, answer is (B)

2. NaI(Tl) scintillation detector is used only for the detection of gamma radiation because
1. it has large scattering cross-section
2. it has small Comption scattering
3. it has small absorption cross-section
4. it has large absorption cross-section

It has large absorption cross-section

Hence, answer is (D)

3. Consider the elastic vibrations of a crystal with two atoms per primitive cell having masses $m$ and $M$ $(M > m)$. Using dispersion relation between the angular frequency $(\omega)$ and wave vector $(K)$, find the frequency for acoustic branch at $K=0$:
1. $\omega=Ka\left(\frac{2C}{m+M}\right)^{1/2}$
2. $\omega=2Ka\left(\frac{2C}{m+M}\right)^{1/2}$
3. $\omega=\frac{1}{2}Ka\left(\frac{2C}{m+M}\right)^{1/2}$
4. $\omega=\frac{1}{2}Ka\left(\frac{2C}{M}\right)^{1/2}$

The dispersion relation is given by $${\scriptstyle m M \omega^4-2C (m +M)\omega^2+2C^2(1−\cos{Ka})=0}$$ For $K\rightarrow0$, $\cos{Ka}=1-\frac{1}{2}K^2a^2$ $${\scriptstyle m M \omega^4-2C (m +M)\omega^2+C^2K^2a^2=0}$$ $${\textstyle \omega^2=2C\left(\frac{1}{m}+\frac{1}{M}\right)}\quad\text{Optical branch}$$ $$\omega^2=\frac{C^2K^2a^2}{2(m+M)}\quad\text{Acoustic branch}$$ $$\omega=\frac{1}{2}Ka\left(\frac{2C}{m+M}\right)^{1/2}$$

Hence, answer is (C)

4. In nuclear direct reactions, time of interaction is of the order of :
1. $10^{-10}$ sec
2. $10^{-16}$ sec
3. $10^{-22}$ sec
4. $10^{-30}$ sec

In nuclear direct interaction time must be very short :$10^{-22}$ sec

In compound nuclear interaction time is large :$10^{-18}$ to $10^{-16}$ sec

Hence, answer is (C)

5. The following decay states a conservation law that forbids it because: $$n\rightarrow p+e^-$$
1. conservation of angular momentum and conservation of Lepton numbers are both violated
2. conservation of baryon number and conservation of Lepton number are both violated
3. conservation of energy is violated
4. conservation of electric charge is violated

$$n\rightarrow p+e^-$$

1. There is conversation of electric charge as protron has charge $+e$, electron has charge $-e$ and neutron is charge-less.
2. baryon number is conserved since baryon number of proton is $+1$, of neutron is $+1$ and of electron is $0$
3. conservation of Lepton number is violated since Lepton number of proton is $0$, of neutron is $0$ and of electron is $1$
4. conservation of angular momentum is also violated since, when the decay of the neutron into a proton and an electron was observed, it did not fit the pattern of two-particle decay. That is, the electron emitted does not have a definite energy as is required by conservation of energy and momentum for a two-body decay.

Hence, answer is (A)