The electron in a given Bohr orbit `E_(n) = - 1.54 eV`. Calculate (i) its kinetic energy (ii) potential energy and (iii) wavelenght of light emitted , when the electron makes a transition to the ground state . Ground state energy is ` - 13.6e V`

Explanation

(i) As total energy in given Bohr orbit `E_(n) = - 1.5 eV`
`therefore ` Kinetic energy of eletron `K_(n) = - E_(n) = + 1.5 eV`
(ii) The potential energy of electron `U_(n) = 2 E_(n) = - 3.0 eV`
(iii) If electron makes a transition from ` - 1.5 eV` state to ground state having energy ` - 13.6 eV` , the energy of radiated photon.
`E = - 1.5 -(-13.6) eV = 12.1 eC = 12.1 xx 1.60 xx 10^(19)J`
As `E =hv = (hv)/(lambda) `, hence wavelenght of light emitted is equal to
`lambda = (hc)/(E) = (6.63xx 10^(-34) xx 3 xx 10^(8))/(12.1 xx 1.60 xx 10^(-19)) = 1.027 xx 10^(-7) m = 102.7` nm.
`lambda =(hc)/(E) = (6.63 xx 10^(-34) xx 3 xx 10^(8))/(12.1 xx 1.60 xx 10^(-19)) = 1.027 xx 10^(-7) m = 102.7 nm`