Question

The critical frequency of emitting photo-electrons from a metal surface is $5\times {10}^{14}se{c}^{-1}$. Calculate the kinetic energy of photo-electron emitted by exposing the surface with the radiations of frequency ${10}^{15}se{c}^{-1}$. What should be the frequency of radiation to produce photoelectrons having twice the kinetic energy of those produced by the radiation of same frequency?

• A. $KE=3.31\times {10}^{-17}J,v=15\times {10}^{13}se{c}^{-1}$
• B. $KE=3.31\times {10}^{-19}J,v=15\times {10}^{14}se{c}^{-1}$
• C. $KE=10\times {10}^{-19}J,v=22\times {10}^{14}se{c}^{-1}$
• D. $KE=8.28\times {10}^{-19}J,v=19\times {10}^{10}se{c}^{-1}$

Answer 1

The correct option is B $KE=3.31\times {10}^{-19}J,v=15\times {10}^{14}se{c}^{-1}$

Critical frequency $=5\times {10}^{14}se{c}^{-1}$
On exposing the surface to frequency of ${10}^{15}se{c}^{-1}=10\times {10}^{14}$
$hv=h{v}_0+KE$
$h\times 10\times {10}^{14}=h\times 5\times {10}^{14}+KE$
$KE=10\times {10}^{14}-5\times {10}^{14}\times h$
$=5\times {10}^{14}\times 6.626\times {10}^{-34}=3.31\times {10}^{-19}J$
If $KE$ is twice of this value, i.e., $10\times {10}^{14}h$, then
$hv=5\times {10}^{14}\times h+10\times {10}^{14}h$
$\therefore v=15\times {10}^{14}se{c}^{-1}$