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The de Brogile wavelength of an electron in the 4^th Bohr orbit is:
Option: 1
$6\pi a_{0}$

Option: 2
$2\pi a_{0}$

Option: 3
$8\pi a_{0}$

Option: 4
$4\pi a_{0}$

Answers (1)

best_answer

We have:

n $\mathrm{\lambda }$ = $\mathrm{2\pi r_{n}}$

n = 4

$\mathrm{2\pi \frac{n^{2}}{z}a_{o}\: =\: n\lambda }$

$\mathrm{2\pi \frac{4^{2}}{1}a_{o}\: =\: n\lambda }$

$\mathrm{\lambda \: =\: 8\pi a_{o}}$

Therefore, Option(3) is correct

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