(a) According to Bohr's postulates, in a hydrogen atom, a single electron revolves around a nucleus of charge+e. For an electron moving with a uniform speed in a circular orbit on a given radius, the centripetal force is provided by the coulomb force of attraction between the electron and the nucleus.
Therefore,
mv2r=1(Ze)(e)4πϵ0r2
⟹ mv2=14πϵ0Ze2r ....(1)
So,
Kinetic Energy, K.E=12mv2
K.E=14πϵ0Ze2r
Potential Energy is given by, P.E=14πϵ0(Ze)(−e)r
Therefore total energy is given by, E=K.E+P.E
E=14πϵ0Ze22r+(−14πϵ0Ze2r)
E=−14πϵ0Ze22r
For, nth orbit, E can be written as En,
En=−14πϵ0Ze22rn ......(2)
Now, using Bohr's postulate for quantization of angular momentum, we have
mvr=nh2π
⟹ v=nh2πmr
Putting the values of v in equation (1), we get
mr (nh2πmr)2=14πϵ0Ze2r2
=−mZ2e48ϵ0h2n2
⟹ ϵ0h2n2πmZe2
Now putting a]value of rn in equation (2), we get
En=−14πϵ0Ze22(ϵ0h2n2πm)
=−mZ2e48ϵ0h2n2
⟹ En=−Z2Rhcn2, where R=me48ϵ20ch3
where R is the Rydberg constant.
(b) Negative sign shows that electron remain bound with the nucleus.
(c) If electron jumps from ni=4,5,6 to nf=3, the energy of the line spectra
△E=me48ϵ20h2(1n2f−1n2i)