The ratio of wavelengths of electron waves in two orbits is 4:3. Then the ratio of kinetic energies of electrons in those orbits is

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16 :9

4 :3

9 : 16

3 : 4

answer is A.

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Detailed Solution

$\begin{aligned} λ = \frac{h}{mv} \\ \Rightarrow λ^{2} = \frac{h^{2}}{m^{2} v^{2}} \\ \Rightarrow {mv}^{2} = \frac{h^{2}}{{mλ}^{2}} \\ ∵ KE (K) = \frac{1}{2} {mv}^{2} \\ ∴ KE (K) = \frac{1 h^{2}}{2 {mλ}^{2}} \end{aligned}$

$\begin{aligned} \Rightarrow \frac{K_{1}}{K_{2}} = {(\frac{λ_{2}}{λ_{1}})}^{2} = {(\frac{3}{4})}^{2} \\ ∴ K_{1} : K_{2} = 9 : 16 . \end{aligned}$

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