The de Broglie wavelength of an electron in the

Question:

The de Broglie wavelength of an electron in the $4^{\text {th }}$ Bohr orbit is:

  1. $2 \pi \mathrm{a}_{0}$

  2. $4 \pi \mathrm{a}_{0}$

  3. $6 \pi \mathrm{a}_{0}$

  4. $8 \pi \mathrm{a}_{0}$


Correct Option: , 4

Solution:

$2 \pi r=n \lambda$

$r=\frac{n^{2} a_{0}}{Z}$

$2 \pi \times \frac{4^{2}}{1} a_{0}=4 \lambda$

$\lambda=2 \pi \times \frac{4}{1} a_{0}$

$\lambda=8 \pi a_{0}$

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