Practice Start Test

Atoms and Nuclei

NEET Physics · 119 questions · Page 11 of 12 · Click an option or "Show Solution" to reveal answer

Q101
Solar energy is mainly caused due to
A burning of hydrogen in the oxygen
B fission of uranium present in the Sun
C fusion of protons during synthesis of heavier elements
D gravitational contraction
Correct Answer
Option C
Solution

As a result of fusion, enormous amount of heat is liberated which is the main cause of source of solar energy.

Q102
An electron is moving round the nucleus of a hydrogen atom in a circular orbit of radius r. The Coulomb force F→\overrightarrow F between the two is
A Ke2r2r^K{{{e^2}} \over {{r^2}}}\widehat r
B −Ke2r3r^ - K{{{e^2}} \over {{r^3}}}\widehat r
C Ke2r3r^K{{{e^2}} \over {{r^3}}}\widehat r
D −Ke2r2r^ - K{{{e^2}} \over {{r^2}}}\widehat r
Correct Answer
Option D
Solution

The charge on hydrogen nucleus q 1 = +e charge on electron, q 2 = –e Coulomb force, F =

K(+e)(−e)r2K{{\left( { + e} \right)\left( { - e} \right)} \over {{r^2}}}

=

−Ke2r3r→- {{K{e^2}} \over {{r^3}}}\overrightarrow r

=

−Ke2r2r^- {{K{e^2}} \over {{r^2}}}\widehat r
Q103
A nuclear reaction given by Z X A → \to z+1 Y A + −-1 e 0 + v‾\overline v represents
A β\beta -decay
B γ\gamma -decay
C fusion
D fission
Correct Answer
Option A
Solution

It represents β\beta-decay. As Emission of electron (e – ) + antineutrino (

ν‾\overline \nu

) ⇒\Rightarrow β\beta-decay.

Q104
In which of the following systems will the radius of the first orbit (n = 1) be minimum ?
A doubly ionized lithium
B singly ionized helium
C deuterium atom
D hydrogen atom
Correct Answer
Option A
Solution

As Radius of first orbit, r ∝\propto

1Z{1 \over Z}

for doubly ionized lithium Z (= 3) will be maximum, hence for doubly ionized lithium, r will be minimum.

Q105
A deutron is bombarded on 8 O 16 nucleus then α\alpha -particle is emitted. The product nucleus is
A 7 N 13
B 5 B 10
C 4 Be 9
D 7 N 14
Correct Answer
Option D
Solution

8 O 16 + 1 H 2 →\to 7 N 14 + 2 He 4 So when a deuteron is bombarded on 8 O 16 nucleus then an α\alpha-particle ( 2 He 4 ) is emitted and the product nucleus is 7 N 14 .

Q106
A sample of radioactive element containing 4 × \times 10 16 active nuclei. Half life of element is 10 days, then number of decayed nuclei after 30 days
A 0.5 × \times 10 16
B 2 × \times 10 16
C 3.5 × \times 10 16
D 1 × \times 10 16
Correct Answer
Option C
Solution

N = 4 ×\times 10 16

(12)3010{\left( {{1 \over 2}} \right)^{{{30} \over {10}}}}

=

12{1 \over 2}

×\times 10 16 ∴\therefore Atoms decayed = 4 ×\times 10 16 -

12{1 \over 2}

×\times 10 16 = 3.5 ×\times 10 16

Q107
Which of the following are suitable for the fusion process ?
A light nuclei
B heavy nuclei
C element lying in the middle of the periodic table
D middle elements, which are lying on binding energy curve.
Correct Answer
Option A
Solution

The nuclei of light elements have a lower binding energy than that for the elements of intermediate mass.

They are therefore less stable; consequently the fusion of the light elements results in more stable nucleus.

Q108
Energy released in nuclear fission is due to
A some mass is converted into energy
B total binding energy of fragments is more than the binding energy of parantal element
C total binding energy of fragments is less than the binding energy of parental element
D total binding energy of fragments is equal to the binding energy of parental element.
Correct Answer
Option A
Solution

Energy released in nuclear fission is due to some mass is converted into energy.

Q109
M n and M p represent the mass of neutron and proton respectively. An element having mass M has N neutrons and Z protons, then the correct relation will be
A M < {N ⋅ \cdot M n + Z ⋅ \cdot M p }
B M > {N ⋅ \cdot M n + Z ⋅ \cdot M p }
C M = {N ⋅ \cdot M n + Z ⋅ \cdot M p }
D M = N {M n + M p }
Correct Answer
Option A
Solution

Given : Mass of neutron = M n Mass of proton = M p ; Atomic mass of the element = M ; Number of neutrons in the element = N and number of protons in the element = Z.

We know that the atomic mass (M) of any stable nucleus is always less than the sum of the masses of the constituent particles.

Therefore, M << [NM n + ZM p ].

X is a neutrino, when b-particle is emitted.

Q110
The interplanar distance in a crystal is 2.8 × \times 10 −-8 m. The value of maximum wavelength which can be diffracted
A 2.8 × \times 10 −-8 m
B 5.6 × \times 10 −-8 m
C 1.4 × \times 10 −-8 m
D 7.6 × \times 10 −-8 m
Correct Answer
Option B
Solution

2dsinθ\theta = nλ\lambda ∵\because

−1≤sin⁡θ≤1- 1 \le \sin \theta \le 1

Therefore λ\lambda max = 2d ⇒\Rightarrow λ\lambda max = 2 ×\times 2.8 ×\times 10 −-8 = 5.6 ×\times 10 −-8 m

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