Question

Light of two different frequency whose photon have energies 1 eV and 2.5 eV respectively, successively illuminate a metal of work function 0.5eV. The ratio of maximum speed of the emitted electron will be:

• A. 1 : 1
• B. 1 : 2
• C. 1 : 3
• D. 1 : 5

Answer 1

The correct option is B 1 : 2

$h{v}_1=1eV,h{v}_2=2.5eV$
$h{v}_0=W=0.5eV$
We have, $\frac{1}{2}m{v}^2=hv-W=hv-h{v}_0$
$\therefore \frac{v_1^2}{v_2^2}=\frac{h{v}_1-h{v}_0}{h{v}_2-h{v}_0}$
$=\frac{1-0.5}{2.5-0.5}=\frac{0.5}{2}=\frac{5}{20}=\frac{1}{4}$
Ratio of $v_1$ and $v_2$ = 1 : 2.