Light of wavelength 488 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 V. Find the work function of the material from which the emitter is made.

Solution

Given,
Wavelength of light, λ = 488 nm = 488 x 10^–9 m
Stopping potential, V₀ = 0.38V

As, ${eV}_{0} = hv - ϕ_{0}$

$\Rightarrow$ ${eV}_{0} = h \frac{c}{λ} - ϕ_{0}$

$\Rightarrow$ $ϕ_{0} = \frac{hc}{λ} - {eV}_{0}$

$= (\frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{488 \times 10^{- 9} \times 1.6 \times 10^{- 19}} - \frac{1.6 \times 10^{- 19} \times 0.38}{1.6 \times 10^{- 19}})$

$= (2.55 - 0.38) eV = 2.17 eV .$

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Wave Optics

Light of wavelength 488 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 V. Find the work function of the material from which the emitter is made.

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