33 Years NEET-AIPMT Chapterwise Solutions - Physics 2020

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Dual Nature of Radiation and Matter
107
2
l p ∝ l e
73. (d) : de Broglie wavelength of neutrons in thermal
equilibrium at temperature T is
h
λ=
, where m is the mass of the neutron
2mk B T
Here, m = 1.67 × 10 –27 kg, k B = 1.38 × 10 –23 J K –1
h = 6.63 × 10 –34 J s
−34
663 . × 10
∴ λ =
−27 −23
2× 1. 67 × 10 × 138 . × 10 × T
−34 25 −10
308 . × 10 × 10 308 . × 10 30.
8
=
=
m=
Å
T T T
74. (d) : Radius of the circular path of a charged particle
in a magnetic field is given by
R = mv or mv = RBq
Bq
Here, R = 0.83 cm = 0.83 × 10 –2 m, B = 0.25 Wb m –2
q = 2e = 2 × 1.6 × 10 –19 C
\ mv = (0.83 × 10 –2 )(0.25)(2 × 1.6 × 10 –19 )
−34
h
66 . × 10
de Broglie wavelength,
Planck’s constant (h) = 6.6 × 10
λ= =
= 001 . Å
–34 J s
mv
−2 −19
083 . × 10 × 0. 25 × 2× 1.
6×
Energy 10 of an electron (E) = 100 × (1.6 × 10 –19 ) J
−34
66 . × 10
−34
−12
h
66 . × 10
=
× 10 m = 001 . Å
−2 −19
or λ= =
083 . × 10 × 0. 25 × 2× 1.
6×
10
2mE
−31 −19
2× 9. 1× 10 × 100 × 16 . × 10
75. (a) : de Broglie wavelength associated with an = 1.2 × 10 –10 m = 1.2 Å
electron is
83. (a) : The kinetic energy of an electron
h
h
λ = or P =
1 2
P λ
× mv = eV
2
∆P
∆λ
∴ =−
P λ ; P
= 05 .
or final velocity of electron (v) = 2eV
P initial 100
m
84. (d) : Momentum of electrons,
P initial = 200P
(p
76. (*) : The de Broglie wavelength l associated with
e ) = 2meV
the electrons is
λ= 1 227 . V nm
where V is the accelerating potential in volts.
or λ∝ 1 V
3
λ1
V2
100 × 10
λ1
∴ = = = 2 or λ =
λ
3 2
2 V1
25 × 10
2
*None of the given options is correct.
h
h
77. (c) λ= :
=
−
−
10 6 kg × v 91 . × 10 31 kg× 3×
10
6 m/s
vvv
\ v = 2.7 × 10 –18 m/s
78. (d) : de Broglie wavelength for a particle is given
h h
by λ= = , where m, v and p are the mass, velocity
p mv
and momentum respectively. h is Planck’s constant. Now,
since all the particles are moving with same velocity, the
particle with least mass will have maximum de-Broglie
wavelength. Out of the given four particles (proton,
neutron, a-particles, i.e., He nucleus and b-particles, i.e.,
electrons) b-particle has the lowest mass and therefore it
has maximum wavelength.
79. (a)
80. (a) : K.E. = 1.6 × 10 –19 × 1 J = 1 eV
81. (c) : Potential difference (V) = 100 volts.
Kinetic energy of an electron (K.E.)
= eV = (1.6 × 10 –19 ) × 100 = 1.6 × 10 –17 J
82. (c) : Kinetic energy (E) = 100 eV;
Mass of electron (m) = 9.1 × 10 –31 kg;
1 eV = 1.6 × 10 –19 J and
Momentum for proton (p p ) =
Therefore,
2MeV
λp
h pp
pe
2meV
= = = = ⎛ λe
⎝ ⎜ m ⎞
h p p
⎠
⎟
e p 2MeV
M
⎛ m ⎞
Therefore , λp
= λ
⎝
⎜
M ⎠
⎟
h h
85. (a) : Wavelength (l) = =
mv p . Therefore for same
wavelength of electrons and photons, the momentum
should be same.
h h
86. (a) : de Broglie wavelength, λ= =
p mv
87. (a)