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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).

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#### 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 × 10^{8 }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 E_{k} 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 × 10^{5} ms^{–1}

Hence, the velocity of the photoelectron is 5.84 × 10^{5} ms^{–1}.

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