Modern Physics Ques 209
- Characteristic $X$-rays of frequency $4.2 \times 10^{18} Hz$ are produced when transitions from $L$-shell to $K$-shell take place in a certain target material. Use Mosley’s law to determine the atomic number of the target material. Given Rydberg’s constant $R=1.1 \times 10^{7} m^{-1}$.
(2003, 2M)
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Solution:
Formula:
- Mass defect in the given nuclear reaction:
$$ \begin{aligned} \Delta m & =2(\text { mass of deuterium })-(\text { mass of helium }) \\ & =2(2.0141)-(4.0026)=0.0256 \end{aligned} $$
Therefore, energy is released
$$ \begin{aligned} \Delta E & =(\Delta m)(931.48) MeV=23.85 MeV \\ & =23.85 \times 1.6 \times 10^{-13} J=3.82 \times 10^{-12} J \end{aligned} $$
Efficiency is only $25 %$, therefore,
$$ \begin{alignedat} 25 % \text { of } \Delta E & =\frac{25}{100}\left(3.82 \times 10^{-12}\right) J \\ & =9.55 \times 10^{-13} J \end{aligned} $$
i.e, by the fusion of two deuterium nuclei, $9.55 \times 10^{-13} J$ energy is available to the nuclear reactor.
Total energy required in one day to run the reactor with a given power of $200\ \text{MW}$ :
$$ E _{\text {Total }}=200 \times 10^{6} \times 24 \times 3600=1.728 \times 10^{13} J $$
$\therefore$ Total number of deuterium nuclei required for this purpose
$$ \begin{aligned} n & =\frac{E _{\text {Total }}}{\Delta E / 2}=\frac{2 \times 1.728 \times 10^{13}}{9.55 \times 10^{-13}} \\ & =0.362 \times 10^{26} \end{aligned} $$
$\therefore$ Mass of deuterium required
$=($ Number of $g$-moles of deuterium required $) \times 2 g$
$$ =\frac{0.362 \times 10^{26}}{6.02 \times 10^{23}} \times 2=120.26 \text{ g} $$
BETA
