Question

Light of two different frequencies whose photos have energies 1eV and 2.5eV respectively illuminate a metallic surface whose work function is 0.5eV successively. Ratio of maximum speeds of emissions will be

• A. 1:4
• B. 1:2
• C. 1:1
• D. 1:5

Answer 1

The correct option is B 1:2

Maximum kinetic energy of photo electrons $E=h{\nu }_{incident}-\varphi$
where $\varphi$ is work-function.
Case 1 : $h{\nu }_{incident}=1eV$ and $\varphi =0.5eV$
$⟹E_1=1-0.5=0.5eV$
Case 1 : $h{\nu }_{incident}=2.5eV$ and $\varphi =0.5eV$
$⟹E_2=2.5-0.5=2eV$
Maximum velocity of photo electron $v=\sqrt{\frac{2E}{m}}$
$⟹\frac{v_1}{v_2}=\sqrt{\frac{E_1}{E_2}}=\sqrt{\frac{0.5}{2}}=\frac{1}{2}$