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Question

(a) Using Bohr's postulates, obtain the expression for total energy of the electron in the nth orbit of hydrogen atom.
(b) What is the significance of negative sign in the expression for the energy?
(c) Draw the energy level diagram showing how the line spectra corresponding to Paschen series occur due to transition between energy levels.

Solution
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(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(1n2f1n2i)




892004_492237_ans_3e0e20afeb4647699ae5168fd15332e6.jpg

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