### JEE Main 12 Jan 2019 Evening Question 25

##### Question: If the de Broglie wavelength of the electron in $ n^{th} $ Bohr orbit in a hydrogenic atom is equal to $ 1.5\pi a _0 $ ( $ a _0 $ is Bohr radius), then the value of n/z is [JEE Main Online Paper Held On 12-Jan-2019 Evening]

#### Options:

A) 1.0

B) 1.50

C) 0.75

D) 0.40

## Show Answer

#### Answer:

Correct Answer: C

#### Solution:

- $ 2\pi r _{n}=n\lambda $ $ 2\pi a _0\times \frac{n^{2}}{Z}=n\lambda $ $ 2\pi a _0\times \frac{n}{Z}=1.5\pi a _0 $ $ \frac{n}{Z}=\frac{1.5\pi a _0}{2\pi a _0}=0.75 $