$(a)\;0.692 A^{0}\qquad(b)\;0.548 A^{0}\qquad(c)\;0.312 A^{0}\qquad(d)\;0.212 A^{0}$

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Answer : (b) $\;0.548 A^{0}$

Explanation :

The energy for $\;n^{th}\;$ orbit is given by

$\varepsilon_{n}=-\large\frac{m K^2 Z^2 e^4}{2 n^2 h^2}$

Substituting $\;m=208e_{e}\;, Z=3 \;, K=\large\frac{1}{4 \pi \varepsilon_{0}}$

and $\;\hbar=\large\frac{h}{2 \pi}\;,$ We get

$\varepsilon_{n}=-\large\frac{234 m_{e} e^4}{\varepsilon_{0}^{2} n^2 h^2 } =-1872 \;(\large\frac{M_{e} e^{4}}{8 \varepsilon_{0}^{2} h^3 c})\;\large\frac{h c}{n^2}$

$=-\large\frac{1872R h c}{n^2}$

Where $R=\large\frac{m_{e} e^{4}}{8 \varepsilon_{0}^{2} h^3 c}\;$ is Rydberg const.

where $\;mu\;$-meson jumps from third orbit to first orbit , difference in energy is radiated as a photon of frequency $\; \nu\;$is given by

$h \nu=\varepsilon_{3}-\varepsilon_{1}$

As $\;\nu=\large\frac{c}{\lambda}\;$ we have $\;\large\frac{hc}{\lambda}=\varepsilon_{3}-\varepsilon_{1}$

$=1872 Rhc \;[\large\frac{1}{1^2}-\large\frac{1}{3^2}]$

or $\; \large\frac{1}{\lambda}=1872R\;(1-\large\frac{1}{9}) $

or $\; \lambda=\large\frac{9}{1872 \times 8 \times R}=\large\frac{9}{1872 \times8 \times(1.097 \times10^{7})}=0.5478 \times10^{-10} m$

$=0.5478 A^{0}\;.$

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