Question

Radiation emitted due to transistion of an electron from n = 2 to n = 1 in Hydrogen atom is made to fall on a metal surface with work function 1.2eV. The maximum velocity of photo electrons emitted is nearly equal to

• A. $6\times {10}^{5}m/s$
• B. $3\times {10}^{5}m/s$
• C. $2\times {10}^{5}m/s$
• D. $18\times {10}^{5}m/s$

Answer 1

The correct option is D $18\times {10}^{5}m/s$

$\Delta E=13.6–3.4=10.2eV$
$\Delta E=W+\frac{1}{2}m{v}^2$
$\Delta E–W=\frac{1}{2}m{v}^2$
$1.6\times {10}^{-19}(10.2–1.2)=\frac{1}{2}X9X{10}^{-31}{v}^2$
$\frac{1.6\times 2\times {10}^{12}\times 9}{9}=v^2$
$v^2=32\times {10}^{11}=320\times {10}^{10}$
$v=18\times {10}^{5}m{s}^{-1}$