Question

Radiation of two photon energies twice and five times the work function of metal are incident successively on the metal surface. The ratio of the maximum velocity of photoelectrons emitted is the two cases will be :

• A. $1:2$
• B. $2:1$
• C. $1:4$
• D. $4:1$

Answer 1

The correct option is A $1:2$

As,

Maximum $K.E=$ incident photon energy $-$ work function

$\frac{1}{2}{mV}_1^2=2W-W=W⟶\left(1\right)$

and

$V_e{mV}_2^2=5W-W=4W⟶\left(2\right)$

from $\left(1\right)$ and $\left(2\right)$

$\frac{V_1}{V_2}=\frac{1}{2}$

$V_1:V_2=1:2$