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Question
The threshold wavelength of tungsten is 2.76 x 10-5 cm.
(a) Explain why no photoelectrons are emitted when the wavelength is more than 2.76 x 10-5 cm.
(b) What will be the maximum kinetic energy of electrons ejected in each of the following cases
(i) if ultraviolet radiation of wavelength λ = 1.80 × 10-5 cm and
(ii) radiation of frequency 4 x 1015 Hz is made incident on the tungsten surface?
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Solution
Data: λ0 = 2.76 x 10-5 cm = 2.76 x 10-7 m,
λ = 1.80 × 10-5 cm = 1.80 × 10-7 m,
v = 4 × 1015 Hz,
h = 6.63 × 10-34 J.s,
c = 3 × 108 m/s
(a) For λ > λ0,v < v0 (threshold frequency).
∴ hv < hv0. Hence, no photoelectrons are emitted.
(b) Maximum kinetic energy of electrons ejected
`= "hc" (1/lambda - 1/lambda_0)`
`= (6.63 xx 10^-34)(3 xx 10^8)(10^7/1.8 - 10^7/2.76)`J
= (6.63 × 10-19)(3)(0.5555 - 0.3623)
= (6.63)(3)(0.1932 × 10-19)J
= 3.842 × 10-19J
`= (3.842 xx 10^-19 "J")/(1.6 xx 10^-19 "J"//"eV")`
= 2.40 eV
(c) Maximum kinetic energy of electrons ejected
= hv - `"hc"/lambda_0`
`= (6.63 xx 10^-34)(4 xx 10^15) - ((6.63 xx 10^-34)(3 xx 10^8))/(2.76 xx 10^-7)`
= 26.52 × 10-19 - 7.207 × 10-19
= 19.313 × 10-19 J
= `(19.313 xx 10^19 "J")/(1.6 xx 10^-19 "J"//"eV")`
= 12.07 eV
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