Question

Light of two different frequencies whose photons have energies 1eV and 2.5 eV respectively, successively illuminates a metal of work function 0.5eV. The ratio of maximum kinetic energy of the emitted electron will be

• A. 1 : 5
• B. 1 : 4
• C. 1 : 2
• D. 1 : 1

Answer 1

The correct option is B 1 : 4

By using $E=W_0+K_{max}\Rightarrow K_{max}=E-W_0$
$Hence,K_1=1-0.5=0.5$
$and{K}_2=2.5-0.5=2\Rightarrow \frac{K_1}{K_2}=\frac{1}{4}$