# Illustrations

Example-1

A photon of frequency 5 Ã— 10^{14} Hz is incident on the surface of a metal. If the photoelectric threshold frequency for the metal is 3 Ã— 10^{14} Hz, calculate

- the energy of the photon
- the photoelectric work function in eV.

Solution

- Energy of the photon,
*E*=*hv**h*= 6.625 Ã— 10^{â€“34}J s,*v*= 5 Ã— 10^{14}Hz*E*= (6.625 Ã—10^{â€“34}Ã— 5 Ã— 10^{14})^{â€“19}J = 1 eV*E*= eV - Photoelectric work function
*W*=*hv*_{0}_{= (6.625 Ã— 10-34 Ã— 3 Ã— 1014) J}_{}_{i.e., W = 1.24 eV.}

Example-2

The threshold wavelength for a metal is 680 nm. Calculate the maximum velocity of photoelectrons emitted, when radiation of wavelength 560 nm is incident on the metal.

Solution

Threshold wavelength is Î»_{0} = 680 nm = 680 Ã— 10^{â€“9} m.

We know from Einsteinâ€™s equation

*E* = *W* + *T* (*T* is the kinetic energy of the electrons)

*T* = *E* â€“ *W*

but *E* = *hv* and *W* = *hv*_{0}

âˆ´ *T* = *hv* âˆ’ *hv*_{0} (1)

But , where â€˜*c*â€™ is the velocity of light. Therefore, Eq. (1) becomes

= 6.625 Ã— 10^{â€“34} Ã— 3 Ã— 10^{8} Ã—

= 6.26 Ã— 10^{â€“20} J.