Question

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

Answer 1

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

Absorbed energy=threshold energy+KE

$nv=n{v}_0+KE$

$\therefore n{v}_0=nv-KE$

$6.26\times {10}^{-34}\times v_0=6.26\times {10}^{-34}\times 2\times {10}^{14}-6.63\times {10}^{-20}$

$\therefore v_0=9.999\times {10}^{13}$

$\simeq 1\times {10}^{14}{s}^{-1}$