(C) From the de-Broglie equation, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \times KE}}$.Since $h$ and $m$ are constants for an electron, we have $\l

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(C) From the de-Broglie equation, $\lambda = \frac{h}{p} = \frac{h}{\sqrt{2m \times KE}}$.Since $h$ and $m$ are constants for an electron, we have $\lambda \propto \frac{1}{\sqrt{KE}}$.Therefore, the ratio of wavelengths is $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{KE_2}{KE_1}}$.Given $\frac{KE_1}{KE_2} = \frac{16}{9}$, we get $\frac{\lambda_1}{\lambda_2} = \sqrt{\frac{9}{16}} = \frac{3}{4}$ or $3:4$.

Class 11 Chemistry Structure of Atom De Broglie's principle

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Given the ratio of kinetic energy of an electron in two orbitals is $16:9$. Calculate the ratio of the wavelength of the electron waves.

TS EAMCET 2020 All TS EAMCET

A
$4:3$
B
$9:16$
C
$3:4$
D
$16:9$

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Given the ratio of kinetic energy of an electron in two orbitals is $16:9$. Calculate the ratio of the wavelength of the electron waves.

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