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The wavelengths of electron waves in two orbits is $3: 5 .$ The ratio of kinetic energy of electrons will be
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Verified Answer
The correct answer is:
$25: 9$
According to de-Broglie's equation.
$$
\lambda=\frac{h}{m v} \Rightarrow \lambda^{2}=\frac{h^{2}}{m^{2} v^{2}}
$$
or $m v^{2}=\frac{h^{2}}{m \lambda^{2}}$
$$
\begin{array}{l}
\because \mathrm{KE}(K)=\frac{1}{2} m v^{2} \\
\therefore \mathrm{KE}(K)=\frac{1}{2} \frac{h^{2}}{m \lambda^{2}}
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow \frac{K_{1}}{K_{2}}=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{2}=\left(\frac{5}{3}\right)^{2} \\
\therefore K_{1}: K_{2}=25: 9
\end{array}
$$
$$
\lambda=\frac{h}{m v} \Rightarrow \lambda^{2}=\frac{h^{2}}{m^{2} v^{2}}
$$
or $m v^{2}=\frac{h^{2}}{m \lambda^{2}}$
$$
\begin{array}{l}
\because \mathrm{KE}(K)=\frac{1}{2} m v^{2} \\
\therefore \mathrm{KE}(K)=\frac{1}{2} \frac{h^{2}}{m \lambda^{2}}
\end{array}
$$
$$
\begin{array}{l}
\Rightarrow \frac{K_{1}}{K_{2}}=\left(\frac{\lambda_{2}}{\lambda_{1}}\right)^{2}=\left(\frac{5}{3}\right)^{2} \\
\therefore K_{1}: K_{2}=25: 9
\end{array}
$$
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