Question

Light of two different frequencies whose photons have energies 1 eV and 2.5 eV respectively, successively illuminates a metal of work function 0.5 eV. 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 $K_{max}=E-W_0\Rightarrow \frac{(K_{max}{)}_1}{(K_{max}{)}_2}=\frac{1-0.5}{2.5-0.5}=\frac{0.5}{2}=\frac{1}{4}$.