Modern Physics 5 Question 18
19. The kinetic energy (in $keV$ ) of the alpha particle, when the nucleus ${ } _{84}^{210} Po$ at rest undergoes alpha decay, is
(a) 5319
(b) 5422
(c) 5707
(d) 5818
Passage 2
Scientists are working hard to develop nuclear fusion reactor. Nuclei of heavy hydrogen, ${ } _1^{2} H$ known as deuteron and denoted by $D$ can be thought of as a candidate for fusion reactor. The D-D reaction is ${ } _1^{2} H+{ } _1^{2} H \rightarrow{ } _2^{3} He+n+$ energy. In the core of fusion reactor, a gas of heavy hydrogen is fully ionized into deuteron nuclei and electrons. This collection of ${ } _1^{4} H$ nuclei and electrons is known as plasma. The nuclei move randomly in the reactor core and occasionally come close enough for nuclear fusion to take place.
Usually, the temperatures in the reactor core are too high and no material wall can be used to confine the plasma. Special techniques are used which confine the plasma for a time $t _0$ before the particles fly away from the core. If $n$ is the density (number/volume) of deuterons, the product $n t _0$ is called Lawson number. In one of the criteria, a reactor is termed successful if Lawson number is greater than $5 \times 10^{14} s cm^{-3}$. It may be helpful to use the following : Boltzmann constant $k=8.6 \times 10^{-5} eV / K ; \frac{e^{2}}{4 \pi \varepsilon _0}=1.44 \times 10^{-9} eVm$.
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Answer:
Correct Answer: 19. (A) $\rightarrow$ p, q (B) $\rightarrow p, r$ $(C) \rightarrow p, s \quad$ (D) $\rightarrow p, q, r$
Solution:
- ${ } _{84} Po^{210} \rightarrow{ } _2 He^{4}+{ } _{82} Pb^{206}$
$$ \begin{aligned} \text { Mass defect } \Delta m & =\left(m _{Po}-M _{He}-m _{Pb}\right)=0.005818 u \\ \therefore \quad Q & =(\Delta m)(931.48) MeV=5.4193 MeV \\ & =5419 keV \end{aligned} $$
From conservation of linear momentum,
$$ \begin{aligned} p _{Pb} & =p _{\alpha} \\ \therefore \quad \sqrt{2 m _{Pb} k _{Pb}} & =\sqrt{2 m _{\alpha} k _{\alpha}} \text { or } \frac{k _{\alpha}}{k _{Pb}}=\frac{m _{Pb}}{m _{\alpha}}=\frac{206}{4} \\ \therefore \quad k _{\alpha} & =\frac{206}{206+4}\left(k _{\text {total }}\right) \\ & =\frac{206}{210}(5419)=5316 keV \end{aligned} $$
