JEE Main Physics Atoms And Nuclei Nuclear Exercise by ednexa.com

1. Atoms Molecules and Nuclei
Exercise
1. State the postulates of Bohr as theory of hydrogen atom.
Ans: Postulate 1 : The electron in a hydrogen atom revolves in circular orbit around
the nucleus with nucleus as the center of the orbit. The necessary centripetal
force for circular motion is provided by electrostatic force of the attraction
between the positively charged nucleus and negatively charged electron.
Centripetal force = electrostatic force of attraction
2
2
0
2
.
4
1
r
e
r
mv


Postulate 2 : The electron revolves around the nucleus
only in those orbits for which the angular momentum is
equal to an integral multiple of
2
h
, where h is Plank’s
Constant,
Angular momentum = mvr =
2
nh
…………..(2)
n is principal quantum number.
Postulate 3 : When electron jumps from orbit of higher
energy to an orbit of lower energy, it radiates energy in the
form of photon. The energy of emitted photon is equal to
the difference between the energies of two orbits in which
transition is taking place.
Let the initial higher energy orbit has principle quantum number = n,
Let final orbit on which the electron jumps has principle quantum number = P
En –Ep = hν
ν = frequency of emitted radiation.
2. Obtain the expression for radius of nth
Bohr orbit and show that the radius
is proportional to square of the principle quantum number.
Ans:
i. Consider the electron of hydrogen atom revolving around the nucleus in a
circular orbit of radius r, with linear velocity v.
ii. According to Bohr’s first postulate – 2
2
0
2
.
4
1
r
e
r
mv


M = mass of electron
R = radius of Bohr’s orbit
e = charge on electron

2. ε0= permittivity of free space
r
e
mv
2
0
2
.
4
1


∴
mr
e
v
0
2
2
4

…………..(i)
iii. According to Bohr’s second postulate –
mvr =
2
nh
v = linear velocity of electron
h = plank’s constant
n = principal quantum number
v =
mr
nh
2
∴ v2
= 222
22
2 rm
hn
 …………..(ii)
iv. Form equation no. 1 and 2
222
22
0
2
44 rm
hn
mr
e


∴
mr
hne

22
0
2

∴ πmre2
= ε0n2
h2
∴ 2
22
0
me
hn
r



This is the expression for radius of Bohr’s orbit
As ε0, h2
, π m and e are constant
∴ r n2
i.e. radius of Bohr’s orbit of hydrogen atom is directly proportional to
square of principal quantum number.

3. 3. Show that angular speed of electron in nth
Bohr orbit is equal to –
ω =
4
2 3 3
0
me
2 h n


or frequency of revolution, f =
4
2 3 3
0
me
4 h n
Ans: The linear speed of electron in nth
Bohr orbit is given by
nh
e
0
2
2
 
i. We know that
v = rω
r
v
=
Putting the values of v and r
2 2
2 2
0 0
e me
2 nh n h

  
 
4
2 3 3
0
me
2 n h

 

This is the expression for angular velocity of an electron in the nth
Bohr
orbit.
As π, m, e, 2, ε0 and h are constant.
4
2 3 3
0
me
2 n h

 3
1
n
 
ii. ω = 2πγ
γ = frequency of revolution



2

Putting the value of ω from equation no. (1)
4
2 3 3
0
me
2 n h

 

4
2 3 3
0
me
4 n h
 

This is expression for frequency of revolution of an electron in the nth
Bohr orbit
As m, e, 4, ε0 and h are constant

4. ∴ 3
1
n
 
iii. Period of revolution =
1
frequencyof revolution
i.e. T =
1

Putting the value of γ
2 3 3
0
4
4 n h
T
me


This is the expression for period of revolution of an electron in the nth
Bohr
orbit
As 4, ε0, h, m and e are constant
∴ T∝ n3
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