First slide

Wave nature of matter

Question

The circumference of the second Bohr orbit of electron in hydrogen atom is 600 nm. The potential difference that must be applied between the plates so that the electrons have the de Broglie wavelength corresponding in this circumference is

Moderate

A

$10^{- 5} V$
B

$\frac{5}{3} \times 10^{- 5} V$
C

$5 \times 10^{- 5} V$
D

$3 \times 10^{- 5} V$

Solution

de Broglie wavelength of electron in hydrogen atom $= \frac{h}{m v} = \frac{2 π r_{n}}{n}$
For second Bohr orbit, $λ = \frac{600 \times 10^{- 9}}{2} = 3000 \times 10^{- 9} m$
$\begin{array}{rcl} λ & = & \sqrt{\frac{150}{V}} Å = 300 Å \\ ∴ V & = & \frac{150}{(3000)^{2}} = \frac{5}{3} \times 10^{- 5} V \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App