Question

The maximum kinetic energy of the photoelectrons ejected from a metal when it is irradiated with a radiation of frequency $2\times {10}^{14}{s}^{-1}$ is $6.63\times {10}^{-20}J$. The thereshold frequency of the metal is:

• A. $2\times {10}^{14}{s}^{-1}$
• B. $3\times {10}^{14}{s}^{-1}$
• C. $2\times {10}^{-14}{s}^{-1}$
• D. $1\times {10}^{14}{s}^{-1}$

Answer 1

The correct option is D $1\times {10}^{14}{s}^{-1}$

$hv=h{v}_o+KE$
$h{v}_o=hv-KE6.626\times {10}^{-34}\times v_o=6.626\times {10}^{-34}\times 2\times {10}^{14}-6.63\times {10}^{-20}{v}_o=\frac{1.3252\times {10}^{-19}-6.63\times {10}^{-20}}{6.626\times {10}^{-34}}=9.99\times {10}^{13}={10}^{14}{s}^{-1}$