A photon of wavelength ` 4xx 10^(-7)` m strikes on metal surface. The work funcation of the metal being 2.13eV. Calculate (i) the energy of the photon (eV) . (ii) the kinetic energy of the emission, and (ii) the velcoity of the photoelectron `(1 eV= 1.6020 xx 10^(-19) J )`

Answer

`0.974eV, 5.85 xx 10^(5)ms^(-1)`

Explanation

According to Einstein.s photoelectric equation
`hv=W+1/2mv^2`
In the present case,
`v=c/lambda=(3 xx 10^(8))/(4 xx 10^(-7)) =7.5 xx 10^(14)Hz`
`h=6.626 xx 10^(-34) Js`
`W=2.13eV`
`=2.13 xx 1.602 xx 10^(-19) xx 10^(-19) (therefore 1eV=1.602 xx 10^(-19)J))`
Kinetic energy of photon `=1/2 mv^2=hv-W`
`=6.626 xx 10^(-34) xx 7.5 xx 10^(14)-3.14xx 10^(-19)`
`=1.56 xx 10^(-19)J =0.974eV`
Morever, Kinetic energy `=1/2mv^2`
`or v=((2 xx "Kinetic energy")/m)^1/2`
`=(2 xx 1.56 xx 10^(-19))/(9.108 xx 10^(-31))^(1/2)`
`("mass of electron "=9.108 xx 10^(-34)kg)`
`=5.85 xx 10^(5)ms^(-1)`.