Problem set 40

1. If superconducting lead has critical temperature of 7.26 K at zero magnetic field and a critical field of $8\times10^5 A/m$ at 0K, the critical field at 5K is :
1. $6.3\times 10^5 A/m$
2. $2.5\times 10^5 A/m$
3. $4.2\times 10^5 A/m$
4. $1.5\times 10^5 A/m$
$H_{c_0}=8\times10^5A/M$ critical field at 0 K
$T=5K$ Temperature at which critical field of lead is to be found outside
$T_c=7.26 K$ Critical temperature of lead
\begin{align*} H_{c_T}&=H_{c_0}\left(1-\left(\frac{T}{T_c}\right)^2\right)\\& =8\times10^5\left(1-\left(\frac{5}{7.26}\right)^2\right)\\ &=4.2\times10^5 A/m \end{align*}
2. The ratio of the sizes (in terms of radii) of $^{208}_{82}Pb$ and $^{26}_{12}Mg$ nuclei is approximately:
1. 2
2. 4
3. $2\sqrt{2}$
4. 8
Nuclear radius is given by $R=R_0A^{1/3}$ $$R_{Pb}=R_0A_{Pb}^{1/3}\quad R_{Mg}=R_0A_{Mg}^{1/3}$$ \begin{align*} \frac{R_{Pb}}{R_{Mg}}&=\left(\frac{A_{Pb}}{A_{Mg}}\right)^{1/3}\\ &=\left(\frac{208}{26}\right)^{1/3}=2 \end{align*} Hence, answer is (A)
3. What is the energy of a gamma radiation backscattered at an angle $180^o$, if the incident energy is 10 MeV?
1. 10 MeV
2. 5 MeV
3. 0.511 MeV
4. 0.25 MeV
If a photon with energy $E_0$ strikes a stationary electron, then the energy of the scattered photon, $E$, depends on the scattering angle, $\theta$, that it makes with the direction of the incident photon according to the following equation: $$\left[\frac{1}{E}-\frac{1}{E_0}\right]=\frac{1}{m_ec^2}(1-\cos\theta)$$ $m_e c^2 = 511 keV = 0.511 MeV$ \begin{align*}\left[\frac{1}{E}-\frac{1}{10}\right]&=\frac{1}{0.511}(1-\cos180)\\ &=3.913894325 \end{align*} \begin{align*} \frac{1}{E}&=3.913894325-0.1\\ &=4.02213894325 \end{align*} $$E=0.25 MeV$$ Hence, answer is (D)
4. Ionisation chamber is effectively used for the measurement of :
1. Radiation
2. Radiation Dose
3. strength of radiation
4. Energy of radiation
Ionisation chamber is effectively used for the measurement of Radiation Dose as it provides an output that is proportional to dose.
Hence, answer is (B)
5. A satisfactory quenching gas in G.M. tube must have the following property:
1. Ionisation potential should be equal to the main counting gas in the tube
2. Ionisation potential should be higher than that of main counting gas in the tube
3. It must have very narrow ultraviolet absorption bands
4. When in excited state it must prefer to dissociate rather than to de-excite by the emission of photon
Quenching gas must have following properties:
1: Its ionisation potential should be less than the main gas
2: It should be complex and prone to dissociation rather than de-excite by photon emission
3: It should have sharp and intense absorption band in UV so that it quickly dissociates by absorbing undesired UV causing spurious discharge.