The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface 5 V , the incident radiation lies in :

X ray region

No worries! We‘ve got your back. Try BYJU‘S free classes today!

ultra violet region

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

infra red region

No worries! We‘ve got your back. Try BYJU‘S free classes today!

visible region

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B ultra violet region
Given,

$V_{0} = 5 V$

$\frac{h c}{λ} = 6.2 e V$

From Einstein formula,

$W = \frac{h c}{λ} - e V_{0} = 6.2 e V - 5 e V$

$W = 1.2 e V$

$\frac{h c}{λ_{i}} = 1.2 e V$

$λ_{i} = \frac{12400 e V}{1.2 e V} = 10333 \times 10^{- 10} m$

$λ_{i} = 1033.3 n m$

The wavelength of incident wave comes in the region of ultra violet.

The correct option is B.

Suggest Corrections

Similar questions

6.2 e V

5 V

6.2 e V

5 V l i e s

3.3 e V

2.4 e V

Join BYJU'S Learning Program