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Question
Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is −1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity (∼105 W m−2) red light of wavelength 6328 Å produced by a He-Ne laser?
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Solution
Wavelength of ultraviolet light, λ = 2271 Å = 2271 × 10−10 m
Stopping potential of the metal, V0 = 1.3 V
Planck’s constant, h = 6.6 × 10−34 J
Charge on an electron, e = 1.6 × 10−19 C
Work function of the metal = `phi_0`
Frequency of light = v
We have the photo-energy relation from the photoelectric effect as:
`phi_0` = hv − eV0
= `"hc"/lambda - "eV"_0`
= `(6.6 xx 10^(-34) xx 3 xx 10^8)/(2271 xx 10^(-10)) - 1.6 xx 10^(-19) xx 1.3`
= `8.72 xx 10^(-19) - 2.08 xx 10^(-19)`
= `6.64 xx 10^(-19) "J"`
= `(6.64 xx 10^(-19))/(1.6 xx 10^(-19))`
= 4.15 eV
Let v0 be the threshold frequency of the metal.
∴ `phi_0 = "hv"_0`
`"v"_0 = phi_0/"h"`
= `(6.64 xx 10^(-19))/(6.6 xx 10^(-34))`
= 1.006 × 1015 Hz
Wavelength of red light, `lambda_"r"` = 6328 Å = 6328 × 10−10 m
∴ Frequency of red light, `"v"_"r" = "c"/lambda_"r"`
= `(3 xx 10^8)/(6328 xx 10^(-10))`
= 4.74 × 1014 Hz
Since v0 > vr, the photocell will not respond to the red light produced by the laser.
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