A photon of wavelength ` 4 xx 10^(-7) m` strikes on metal surface , the work function of the metal being ` 2. 13 eV` Calculate :
(i) the energy of the photon (ev)
(ii) the kinetic energy fo the emission and
the velocity of the photoelectron `( 1 eV = 1 , 6020 xx 10^(-19) J)`,

Explanation

(i) Energy (E) of a photon = hv =`(hc)/lamda`
where
h = Planck’s constant =`6.626xx10^(-34) Js`
c = velocity of light in vacuum =` 3xx 10^(8)m//s`
`lamda` = wavelength of photon =` 4 xx 10 ^(-7) m`
Substituting the values in the given expression of E
`E=(6.626 xx 10^(-34)) (3xx10^(8))/(4xx10^(-7))= 4.9695xx10^(-19) J`
Hence, the energy of the photon is `4.97 xx 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,
` 1/2mv^(2)=hv_(0)`
` Rightarrow 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:
`sqrt((2xx(0.9720xx1.6020xx10^(-19))J)/(9.10939xx10^(31)kg))`
`sqrt(0.3418xx10^(12)m^(2)s^(-2))`
`v=5.48xx10^(5)ms^(-1)`
Hence, the velocity of the photoelectron is `5.84 xx 10^(5)ms^(-1)` .