Question

A metal is irradiated with light of frequency $3.2\times {10}^{16}Hz$. The ejected photoelectron has kinetic energy $\frac{3}{4}$th of the energy of the absorbed photon. The threshold frequency of the metal would be:

• A. $1.6\times {10}^{15}Hz$
• B. $8\times {10}^{15}Hz$
• C. $2.4\times {10}^{25}Hz$
• D. $2.4\times {10}^{18}Hz$

Answer 1

The correct option is B $8\times {10}^{15}Hz$

Equation for photoelectric effect

$h\vartheta =h{\vartheta }_0+KE$

where,

$\vartheta \to$ frequency of incident photon.

${\vartheta }_0\to$ Threshold frequency

It is given that $KE=\frac{3}{4}h\vartheta$

$h{\vartheta }_0=\frac{1}{4}h\vartheta$

${\vartheta }_0=\frac{1}{4}\vartheta$

${\vartheta }_0=\frac{1}{4}\ast 3.2\ast {10}^{16}Hz$

${\vartheta }_0=0.8\ast {10}^{16}Hz$

${\vartheta }_0=8\ast {10}^{15}Hz$