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SECTION  A
Objective Type Questions(AlphaParticle Scattering and Rutherford’s Nuclear Model of atom)
1.An alpha particle colliding with one of the electrons in a gold atom loses(1)Most of its momentum(2)About
31
rd of its momentum(3)Little of its energy(4)Most of its energy
Sol.
Answer (3)The mass of an electron is hundred of times lesser than the mass of an alpha particle. Hence the alphaparticles does not transfer much of its energy on collision with the electron.2.According to classical theory, Rutherford atom was(1)Electrostatically stable(2)Electrodynamically unstable(3)Semi stable(4)Stable
Sol.
Answer (1)Rutherford designed his theory to be electrostatically stable.
(Bohr Model of the Hydrogen Atom)
3.The angular momentum of an electron in a hydrogen atom is proportional to (where
r
is radius of orbit)(1)
r
1
(2)
r
1
(3)
r
(4)
r
2
Sol.
Answer (3)The angular momentum of an electron is quantised as
2
nhmvr
=π
While radius of the n
th
orbit is
r
=
n
2
r
0
where,
r
0
= 0.53 Å
Chapter
12Atoms
Solutions
270
AtomsSolutions of Assignment (Set2)
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4.When a hydrogen atom is raised from the ground state to third state(1)Both kinetic energy and potential energy increase(2)Both kinetic energy and potential energy decrease(3)Potential energy increases and kinetic energy decreases(4)Potential energy decreases and kinetic energy increases
Sol.
Answer (3)When hydrogen atom is raised from the ground state to third state
02
3
E E
=
or
0
9
E E
=
orPE is
0
29
E
−
and KE is
0
9
E
while initially, PE was –2
E
0
and KE was
E
0
.Hence, potential energy increases and kinetic energy decreases.5.What is the angular momentum of an electron in Bohr’s hydrogen atom whose energy is –3.4 eV?(1)
h
(2)
h
2
(3)
2
h
(4)
41
Sol.
Answer (1)
02
E E n
= −
2
13.6 eV 3.4
n
=
2
13.63.4
n
=
2
68417
n
= =
or
n
= 2Hence
mvr
=
2
nh
π
or
h
π
6.The energy levels of a certain atom for first, second and third levels are
E
,
34
E
and 2
E
respectively.A photon of wavelength
is emitted for a transition 3
1. What will be the wavelength of emission for transition2
1?(1)
3
(2)
3
(3)
43
(4)
34
Sol.
Answer (2)
E
= 2
E
–
E
=
hc
271
Solutions of Assignment (Set2)Atoms
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E
=
hc
Similarly
2
43
E hc E
2
3
E hc
2
= 3
7.The ground state energy of H  atom is –13.6 eV. The energy needed to ionise H  atom from its second excitedstate is(1)1.51 eV(2)3.4 eV(3)13.6 eV(4)12.1 eV
Sol.
Answer (1)Energy for nth excited state
02
( 1)
E n
=+
To ionise Hatom from its second excited state
13.61.51 eV 9
= =
8.If element with principal quantum number
n
> 4 were not allowed in nature, then the number of possibleelements would be(1)60(2)32(3)4(4)64
Sol.
Answer (1)Number of electron possible in a shell is given by
N
= 2
n
2
where
n
is number of the shell
N
1
= 2
N
2
= 8
N
3
= 18
N
4
= 32Total
N
= 2 + 8 + 18 + 32 = 60 atoms9.The angular speed of electron in the
n
th orbit of hydrogen atom is(1)Directly proportional to
n
2
(2)Directly proportional to
n
(3)Inversely proportional to
n
3
(4)Inversely proportional to
n
Sol.
Answer (3)
mvr
=
2
nh
π
2
2
nhm r
ω =π
2
nr
ω ∝
As,
r
n
2
3
1
n
ω ∝
272
AtomsSolutions of Assignment (Set2)
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10.As the
n
(number of orbit) increases, the difference of energy between the consecutive energy levels(1)Remains the same(2)Increases(3)Decreases(4)Sometimes increases and sometimes decreases
Sol.
Answer (3)The difference in energy, between consecutive energy levels keeps reducing as easily predicted by theequation of the line spectra, for hydrogen atom
2 2
1 1
f i
R c n n
ν = −
11.The magnetic field induction produced at the centre of orbit due to an electron revolving in
n
th
orbit of hydrogenatom is proportional to(1)
n
–3
(2)
n
–5
(3)
n
5
(4)
n
3
Sol.
Answer (2)
23
12 2
nh nhmvr m r n
= → ω = → ω ∝π π
I = q
e
2
I
π
as
r
n
2
0
2
I Br
= µ
0
2 2
Br
µ ω=π
again as
r
n
2
5
1
Bn
∝
12.The speed of an electron in the orbit of hydrogen atom in the ground state is(1)
c
(2)
c
/10(3)
c
/2(4)
c
/137
Sol.
Answer (4)The speed of electron in ground state of an hydrogen atom is about 2.2 × 10
6
m/s. Which is approximately
137
c
, when
c
is the speed of light.13.If the radius of the first orbit of hydrogen atom is 5.29 × 10
–11
m, the radius of the second orbit will be(1)21.16 × 10
–11
metre(2)15.87 × 10
–11
metre(3)10.58 × 10
–11
metre(4)2.64 × 10
–11
metre
Sol.
Answer (1)Radius of
n
th
orbit =
n
2
r
0
Radius of 2
nd
orbit = 4
r
0
[
r
0
5.3 × 10
–11
m]
Hence radius needed
21.16 × 10
–11
m14.The ratio of minimum to maximum wavelength of radiation emitted by transition of an electron to ground stateof Bohr’s hydrogen atom is(1)
43
(2)
41
(3)
81
(4)
83