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Question
A photon of wavelength 4 × 10–7 m strikes on the metal surface, the work function of the metal being 2.13 eV. Calculate
- the energy of the photon (eV),
- the kinetic energy of the emission, and
- the velocity of the photoelectron (1 eV = 1.6020 × 10–19 J).
Solution
(i) Energy (E) of a photon = `"hν" = "hc"/lambda`
Where,
h = Planck’s constant = 6.626 × 10–34 Js
c = velocity of light in vacuum = 3 × 108 m/s
λ = wavelength of photon = 4 × 10–7 m
Substituting the values in the given expression of E:
`"E" = ((6.626xx10^(-34))(3xx10^8))/(4xx10)^(-7)` = `4.9695 xx 10^(-19) "J"`
Hence, the energy of the photon is 4.97 × 10–19 J.
(ii) The kinetic energy of emission Ek is given by
`= "hv" - "hv"_0`
`=("E" - "W")"eV"`
`= ((4.9695xx10^(-19))/(1.6020xx10^(-19))) "eV" - 2.13 " eV"`
= (3.1020 – 2.13) eV
= 0.9720 eV
Hence, the kinetic energy of emission is 0.97 eV.
(iii) The velocity of a photoelectron (ν) can be calculated by the expression,
`1/2 "mv"^2 = "hv" - "hv"_0`
`=> "v" = sqrt((2("hv" - "hv"_0))/"m")`
where (`"hv" - "hv"_0`) is the kinetic energy of emission in Joules and ‘m’ is the mass of the photoelectron. Substituting the values in the given expression of v:
`"v" = sqrt((2xx(0.9720xx1.6020xx10^(-19))"J")/(9.10939xx10^(-31)"kg"`
`= sqrt(0.3418 xx 10^12 "m"^2"s"^(-2))`
v = 5.84 × 105 ms–1
Hence, the velocity of the photoelectron is 5.84 × 105 ms–1.
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