Light of wavelength 5000 A falls on a sensitive plate with work function 1.90 eV. Calculate (a) the energy of the photon in eV, (b) kinetic energy of the emitted photoelectrons and (c) stopping potential.

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Solution

Given,
Planck's constant, h = 6.62 x 10^–34 Js.
(a) energy of a photon,

E = hv

= \frac{h c}{λ} = \frac{6.62 \times 10^{- 34} \times 3 \times 10^{8}}{5000 \times 10^{- 10}} J

i.e.,

E = \frac{6.62 \times 3}{5} \times 10^{- 19} J

= \frac{6.62 \times 3}{5 \times 1.6 \times 10^{- 19}} \times 10^{- 19} eV = 2.48 eV

(b) Kinetic energy of the emitted photoelectrons,

K . E . = E - ϕ_{0} = (2.48 - 1.90) eV = 0.58 eV

{eV}_{0} = K . E .

V_{0} = \frac{K . E .}{e} = \frac{0.58 eV}{e} = 0.58 V .

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