Question

The minimum energy required for the photo-emission of from the surface of metal is $4.95\times {10}^{-19}$ J. Calculate the critical frequency and the corresponding wavelength of the photon required to eject the electron.

Answer 1

Minimum energy required to emit photoelectron is given by,

$E=h{\nu }_0$ where ${\nu }_0$ is the critical frequency

Substituting $E,h$

${\nu }_0=\frac{E}{h}$

${\nu }_0=\frac{4.95\times {10}^{-19}J}{6.63\times {10}^{-34}Js}$

${\nu }_0=7.5\times {10}^{14}Hz$

Now ${\lambda }_0=c{\nu }_0$ is the critical wavelength

${\lambda }_0=\frac{3\times {10}^{8}m/s}{7.5\times {10}^{14}Hz}$

${\lambda }_0=0.4\times {10}^{-6}m=4000A^0$