If the photon of wavelength 150 pm striks an atom and one of its inner bond electron is ejected out with a velocity of 1.5×107m∕s, what is the energy with which it is bond to the nucleus?
Total energy ( E ) of any photon is given by the relation : E=
hc
λ
where, h = Planck's constant =6.6×10−34J−s c= velocity of light =3×10−10m∕s λ= wave-length =1.5×10−10m,i.e, (150pm) Thus, E=
hc
λ
=
6.6×10−34×3×108
1.5×10−10
;E=1.32×10−15J and, energy of ejected electron (E′) E′=
1
2
mv2 where, m= mass of electron =9.1×10−31kg v= velocity of electron =1.5×107m∕s E′=
1
2
mv2=
1
2
×9.1×10−31×(1.5×107)2 E′=1.024×10−16 Thus, total energy of photon = binding energy of electron (B) + energy of ejected electron (E') Thus, 1.32×10−15=(B)+(E) ∴(B)=(E)−(E′)=[1.32×10−15]−[1.024×10−16] =1.2176×10−15J =