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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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$25: 9$
From de-Broglie's equation
$\lambda=\frac{h}{m v}$
$\begin{aligned} \Rightarrow & \lambda^2 & =\frac{h^2}{m^2 v^2} \\ \Rightarrow & m v^2 & =\frac{h^2}{m \lambda^2}\end{aligned}$
$\begin{aligned} & \therefore & \mathrm{KE}(K) & =\frac{1}{2} m v^2 \\ & \therefore & \mathrm{KE}(K) & =\frac{1}{2} \frac{h^2}{m \lambda^2} \\ \Rightarrow & & \frac{K_1}{K_2} & =\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{5}{3}\right)^2 \\ & \therefore & K_1 & : K_2=25: 9\end{aligned}$
$\lambda=\frac{h}{m v}$
$\begin{aligned} \Rightarrow & \lambda^2 & =\frac{h^2}{m^2 v^2} \\ \Rightarrow & m v^2 & =\frac{h^2}{m \lambda^2}\end{aligned}$
$\begin{aligned} & \therefore & \mathrm{KE}(K) & =\frac{1}{2} m v^2 \\ & \therefore & \mathrm{KE}(K) & =\frac{1}{2} \frac{h^2}{m \lambda^2} \\ \Rightarrow & & \frac{K_1}{K_2} & =\left(\frac{\lambda_2}{\lambda_1}\right)^2=\left(\frac{5}{3}\right)^2 \\ & \therefore & K_1 & : K_2=25: 9\end{aligned}$
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