Using the Bohr's postulates, derive the expression for the (a)Speed of the electron
in the nth orbit.(b)radius of the nth orbit of the electron in hydrogen atom. (c)energy
of the electron in the nth orbit.
Answers
Answer:
(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,
r
mv
2
=
4πϵ
0
r
2
1(Ze)(e)
⟹ mv
2
=
4πϵ
0
1
r
Ze
2
....(1)
So,
Kinetic Energy, K.E=
2
1
mv
2
K.E=
4πϵ
0
1
r
Ze
2
Potential Energy is given by, P.E=
4πϵ
0
1
r
(Ze)(−e)
Therefore total energy is given by, E=K.E+P.E
E=
4πϵ
0
1
2r
Ze
2
+(−
4πϵ
0
1
r
Ze
2
)
E=−
4πϵ
0
1
2r
Ze
2
For, n
th
orbit, E can be written as E
n
,
E
n
=−
4πϵ
0
1
2r
n
Ze
2
......(2)
Now, using Bohr's postulate for quantization of angular momentum, we have
mvr=
2π
nh
⟹ v=
2πmr
nh
Putting the values of v in equation (1), we get
r
m
(
2πmr
nh
)
2
=
4πϵ
0
1
r
2
Ze
2
=−
8ϵ
0
h
2
n
2
mZ
2
e
4
⟹
πmZe
2
ϵ
0
h
2
n
2
Now putting a]value of r
n
in equation (2), we get
E
n
=−
4πϵ
0
1
2(
πm
ϵ
0
h
2
n
2
)
Ze
2
=−
8ϵ
0
h
2
n
2
mZ
2
e
4
⟹ E
n
=−
n
2
Z
2
Rhc
, where R=
8ϵ
0
2
ch
3
me
4
where R is the Rydberg constant.
(b) Negative sign shows that electron remain bound with the nucleus.
(c) If electron jumps from n
i
=4,5,6 to n
f
=3, the energy of the line spectra
△E=
8ϵ
0
2
h
2
me
4
(
n
f
2
1
−
n
i
21 )