2.54 If the photon of the wavelength 150 pm strikes an atom and one of it's inner bound electrons are ejected out with a velocity of 1.5 × 107 m s–1, calculate the energy with which it is bound to the nucleus.

The energy of the incident photon (E) is given by,

E=hcλ
=6.626x10-343.0x108ms-1150x10-12m
=1.3252x10-15J
=13.252x10-16J

The energy of the electron ejected (K.E)

=12mv2
=129.10939x10-31kg1.5x107ms-1
= 10.2480 × 10–17 J
= 1.025 × 10–16 J

Hence, the energy with which the electron is bound to the nucleus can be obtained as:

= E – K.E
= 13.252 × 10–16 J – 1.025 × 10–16 J
= 12.227 × 10–16 J

=12.227x10-161.602x10-19eV
=7.6x103eV
5λ0-20004λ0-2000=5.352.552=28.62256.5025
5λ0-20004λ0-2000=4.40177
17.6070λ0-5λ0=8803.537-2000
λ0=6805.53712.607
λ0=539.8nm
λ0=540nm