Question

If the light of frequency $8.2\times {10}^{-14}\text{Hz}$ is incident on the metal, cut off voltage for photo electric emission is $2.03\text{V}$. The threshold frequency falls into which of the following region?

• A. Infrared Region
• B. Ultraviolet Region
• C. Radio waves
• D. X-Rays

Answer 1

The correct option is A Infrared Region

Given :
$\nu =8.2\times {10}^{14}\text{Hz}$
$V_0=2.03\text{V}$

According to Einstein's photoelectric equation,

$e{V}_0=h\nu -h{\nu }_0$

$1.6\times {10}^{-19}\times 2.03=6.6\times {10}^{-34}(8.2\times {10}^{14}-{\nu }_0)$

${\nu }_0=3.3\times {10}^{14}\text{Hz}$

The frequency range of Infra red region is given by

$3\times {10}^{11}-4\times {10}^{14}Hz\to IRRegion.$

Hence, option $\left(A\right)$ is correct.

Why this question? Key note: Keep remember frequency range for all the important rays.