The wavelengths of electron waves in two orbi | vedclass

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(A) From de-Broglie's equation,$\lambda = \frac{h}{mv}$.Since $KE = \frac{1}{2}mv^2 = \frac{p^2}{2m}$ and $p = \frac{h}{\lambda}$,we have $KE = \frac{

Vedclass · Accepted Answer

$25: 9$

Vedclass · Answer

(A) From de-Broglie's equation,$\lambda = \frac{h}{mv}$.Since $KE = \frac{1}{2}mv^2 = \frac{p^2}{2m}$ and $p = \frac{h}{\lambda}$,we have $KE = \frac{h^2}{2m\lambda^2}$.Thus,$KE \propto \frac{1}{\lambda^2}$.Therefore,$\frac{K_1}{K_2} = \left( \frac{\lambda_2}{\lambda_1} \right)^2$.Given $\frac{\lambda_1}{\lambda_2} = \frac{3}{5}$,we have $\frac{\lambda_2}{\lambda_1} = \frac{5}{3}$.So,$\frac{K_1}{K_2} = \left( \frac{5}{3} \right)^2 = \frac{25}{9}$.The ratio of kinetic energy is $25: 9$.

The wavelengths of electron waves in two orbits is $3: 5$. The ratio of kinetic energy of electrons will be

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