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Question

Ultraviolet light of wavelength 2271A from a 100 W mercury source irradiates a photocell made of molybdenum metal. If the stopping potential is 1.3 V, estimate the work function of the metal. How would the photocell respond to a high intensity (105W/m2) red light of wavelength 6328A produced by a He-Ne laser?

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Solution

Step 1: Find the work function.
Formula used: ϕ0=hveV0
Given:
Wavelength of ultraviolet light,
λ=2271A
λ=2271×1010m
Stopping potential of the metal, V0=1.3V
Planck's constant, h=6.6×1034Js
Charge on an electron, e=1.6×1019C
Wavelength of red light,
λR=6328A=6328×1010m
If the work function of the metal is ϕ and the frequency of incident radiation is v,
Then Work function is given by: ϕ0=hveV0
ϕ0=hcλeV0
ϕ0=6.63×1034×3×1082271×10101.6×1019×1.3
ϕ08.75×10192.08×1019
ϕ06.7×1019J
ϕ in electron volt (eV)
=6.7×10191.6×1019eV
ϕ0=4.2 eV
Step 2 : Find the theshold frequency.
Formula used : ϕ0=hv0
Let v0 be the threshold frequency of the metal, then, we have,
ϕ0=hv0
v0=ϕ0h
v0=6.7×10196.63×1034
v0=1.01×1015Hz
Step 3: Find the frequency of red light.
Formula used: vR=cλR
Now frequency of red light
vR=cλR
vR=3×1086328×1010
vR4.74×1014Hz
Since v0>vr, the photo - cell will not respond
howsoever high be the intensity of laser light.
Final Answer : ϕ0=4.2 eV
v01×1015Hz
vR4.74×1014Hz

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