Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • Engineering
  • The de Brogile wavelength of an electron in the 4th Bohr orbit is:Option: 1 <img alt="6\pi a_{0}" src="https://learn.careers360.com/l

The de Brogile wavelength of an electron in the 4th 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

Posted by

mansi

View full answer

Similar Questions