Light of wavelength 488 \mathrm{~nm} is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 \mathrm{~V}. Find the work function of the material from which the emitter is made.
Light of wavelength 488 \mathrm{~nm} is produced by an argon laser which is used in the photoelectric effect. When light from this spectral line is incident on the emitter, the stopping (cut-off) potential of photoelectrons is 0.38 \mathrm{~V}. Find the work function of the material from which the emitter is made.

Wavelength of light produced by the argon laser is given as

\lambda=488 \mathrm{~nm}=488 \times 10^{-9} \mathrm{~m}

Stopping potential of the photoelectrons is given as \mathbf{V}_{\mathbf{0}}=\mathbf{0 . 3 8} \mathbf{~ V}

1 \mathrm{eV}=1.6 \times 10^{-19} \mathrm{~J}

Therefore, \mathrm{V}_{0}=\frac{0.38}{1.6 \times 10^{-19}} \mathrm{eV}

Planck’s constant, h=6.6 \times 10^{-34} \mathrm{~J} \mathrm{~s}

Charge on an electron, \mathbf{e}=\mathbf{1 . 6} \times \mathbf{1 0}^{-19} \mathbf{C}

Speed of light, c=3 \times 10^{8} \mathrm{~m} / \mathrm{s}

Following is the relation for the work function Using Einstein’s photoelectric effect:

\Phi_{o} of the material of the emitter as:

eV_{0} = \frac{h c}{\lambda}- \Phi_{o}

\Phi_{o}=\frac{h c}{\lambda}-\mathrm{eV}_{0}

=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{1.6 \times 10^{-19} \times 488 \times 10^{-9}}-\frac{1.6 \times 10^{-19} \times 0.38}{1.6 \times 10^{-19}}

=2.54-0.38=2.16 \mathrm{eV}

Therefore, the work function of the material with which the emitter is made is 2.16 \mathrm{eV}.