The effective area of a photosensitive material is 4 cm² and the anode is located at a distance of 1.0 m from a mercury arc lamp of 0.10 W, emitting UV radiation of wavelength 2537 Å. The work functio

Question

The effective area of a photosensitive material is 4 cm² and the anode is located at a distance of 1.0 m from a mercury arc lamp of 0.10 W, emitting UV radiation of wavelength 2537 Å. The work function of the material is 2.22 eV. The photon energy, the flux of photons received by the cathode and the cut-off potential, respectively, are

TestHub · Accepted Answer

Given: Power of lamp P = 0.10 W Distance r = 1.0 m Effective area A = 4 cm² =

$4.0 \times 1 0^{- 4}, m^{2}$

Wavelength λ = 2537 Å = 253.7 nm Work function φ = 2.22 eV hc ≈ 1240 eV·nm 1 eV =

$1.602 \times 1 0^{- 19} J$

Photon energy (in eV)

$E_{p h} (e V) = \frac{h c}{λ ( nm )} = \frac{1240}{253.7} = 4.887 e V \approx 4.90 e V$

Photon energy (in J)

$E_{p h} (J) = 4.887 \times 1.602 \times 1 0^{- 19} = 7.83 \times 1 0^{- 19} J$

Power incident on the photosensitive area Fraction of lamp power hitting the area (assuming isotropic emission):

$f = \frac{A}{4 π r ^{2}} = \frac{4.0 \times 1 0 ^{- 4}}{4 π \times 1 ^{2}} = \frac{1.0 \times 1 0 ^{- 4}}{π}$

Power on cathode:

$P_{c} = P \times f = 0.10 \times (\frac{1.0 \times 1 0 ^{- 4}}{π}) = \frac{1.0 \times 1 0 ^{- 5}}{π} = 3.1831 \times 1 0^{- 6} W; (a pp ro x ima t e l y)$

Photon flux (number of photons received per second)

$\dot{N} = \frac{P _{c}}{E _{p h} ( J )} = \frac{3.1831 \times 1 0 ^{- 6}}{7.83 \times 1 0 ^{- 19}} \approx 4.06 \times 1 0^{12} p h o t o n s / s$

Cut-off (stopping) potential

$V_{s} = E_{p h} (e V) - ϕ = 4.887 - 2.22 = 2.667 V \approx 2.68 V$

Answers (photon energy, photon flux, cut-off potential): = 4.90 eV;

$4.06 \times 1 0^{12}$

photons/s; 2.68 V

This corresponds to option (C).

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