Question

Light of two different frequencies whose photons have energies 1eV and 2.5eV successively illuminate a metal of work function 0.5 eV.The ratio of the maximum speeds of the emitted electrons will he

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

Answer 1

Answer: (C)

For maximum speed of the photo electrons $\frac{1}{2}m{v}^2=E_{photon}-\varphi$

where $\varphi =0.5$ eV is the work function of the metal.

Energy of photon $E_1=1eV$

$\therefore$ $\frac{1}{2}m{v}_1^2=1-0.5$

We get $\frac{1}{2}m{v}_1^2=0.5$ eV ....(1)

Energy of photon $E_2=2.5eV$

$\therefore$ $\frac{1}{2}m{v}_2^2=2.5-0.5$

We get $\frac{1}{2}m{v}_2^2=2$ eV ....(2)

Ratio of maximum speeds $\frac{v_1^2}{v_2^2}=\frac{0.5}{2}=\frac{1}{4}$

We get $\frac{v_1}{v_2}=\frac{1}{2}$