Practice Start Test

Atoms and Nuclei

NEET Physics · 119 questions · Page 3 of 12 · Click an option or "Show Solution" to reveal answer

Q21
The half life of a radioactive nuclide is 100 hours. The fraction of original activity that will remain after 150 hours would be :
A 232{2 \over {3\sqrt 2 }}
B 12{1 \over 2}
C 122{1 \over {2\sqrt 2 }}
D 23{2 \over 3}
Correct Answer
Option C
Solution
AA0=(12)t/TH=(12)150/100=122{A \over {{A_0}}} = {\left( {{1 \over 2}} \right)^{t/{T_H}}} = {\left( {{1 \over 2}} \right)^{150/100}} = {1 \over {2\sqrt 2 }}
Q22
A radioactive nucleus ZAX_Z^AX →\to Z−1B{}_{Z - 1}B →\to Z−3C{}_{Z - 3}C →\to Z−2D{}_{Z - 2}D, where Z is the atomic number of element X. The possible decay particles in the sequence are :
A β\beta −- , α\alpha, β\beta +
B α\alpha, β\beta −- , β\beta +
C α\alpha, β\beta + , β\beta −-
D β\beta + , α\alpha, β\beta −-
Correct Answer
Option D
Solution

β\beta + decreases atomic number by 1. α\alpha decreases atomic number by 2. β\beta −- increases atomic number by 1.

Q23
When a uranium isotope 92235U_{92}^{235}U is bombarded with a neutron, it generates 3689Kr_{36}^{89}Kr three neutrons and :
A 4091Zr{}_{40}^{91}Zr
B 36101Kr{}_{36}^{101}Kr
C 36103Kr{}_{36}^{103}Kr
D 56144Ba{}_{56}^{144}Ba
Correct Answer
Option D
Solution
92235U{}_{92}^{235}U

+

01n{}_{0}^{1}n

=

3689U{}_{36}^{89}U

+

01n{}_{0}^{1}n

+

ZAX{}_{Z}^{A}X

Balancing atomic number on both side, we get 92 + 0 = 36 + Z So, Z = 56 Balancing mass number on both side, we get 235 + 1 = 89 + 3 + A So, A = 144 It gives

56144Ba_{56}^{144}Ba
Q24
For which one of the following, Bohr model is not valid?
A Singly ionised helium atom (He + )
B Deuteron atom
C Singly ionised neon atom (Ne + )
D Hydrogen atom
Correct Answer
Option C
Solution

Singly ionized neon has electron count more than one.

Bohr's model is valid for atoms of single electron.

So option (c) is not valid.

Q25
The energy equivalent of 0.5 g of a substance is :
A 4.5×1013J4.5 \times {10^{13}}J
B 1.5×1013J1.5 \times {10^{13}}J
C 0.5×1013J0.5 \times {10^{13}}J
D 4.5×1016J4.5 \times {10^{16}}J
Correct Answer
Option A
Solution
E=mc2=0.5×10−3×(3×108)2=4.5×1013JE = m{c^2} = 0.5 \times {10^{ - 3}} \times {\left( {3 \times {{10}^8}} \right)^2} = 4.5 \times {10^{13}}J
Q26
α\alpha -particale consists of :
A 2 electrons, 2 protons and 2 neutrons
B 2 electrons and 4 protons only
C 2 protons only
D 2 protons and 2 neutrons only
Correct Answer
Option D
Solution

Since, α\alpha-particle is equivalent to He-atom nucleus, hence it has two protons and two neutrons only.

Q27
The total energy of an electron in an atom in an orbit is –3.4 eV. Its kinetic and potential energies are, respectively.
A 3.4 eV, – 6.8 eV
B 3.4 eV, 3.4 eV
C – 3.4 eV, – 3.4 eV
D – 3.4 eV, – 6.8 eV
Correct Answer
Option A
Solution

Apply Bohr's Atomic model for H-atom KE = –T.E and PE = 2T.E Given, T.E = –3.4 eV ∴\therefore KE = +3.4 eV and PE = -6.8 eV

Q28
For a radioactive material, half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is
A 20
B 10
C 30
D 15
Correct Answer
Option A
Solution

Given N 0 = 600, N 1 = 450, T = 10 min Number of nuclei remaining, N = 600 – 450 = 150 after time ‘t’

NN0=150600=14{N \over {{N_0}}} = {{150} \over {600}} = {1 \over 4}

According to the law of radioactive decay, N = N 0 e -λ\lambdat

NN0=(12)tT1/2{N \over {{N_0}}} = {\left( {{1 \over 2}} \right)^{{t \over {{T_{1/2}}}}}}

∴\therefore

150600=(12)tT1/2{{150} \over {600}} = {\left( {{1 \over 2}} \right)^{{t \over {{T_{1/2}}}}}}

⇒\Rightarrow

(12)2=(12)tT1/2{\left( {{1 \over 2}} \right)^2} = {\left( {{1 \over 2}} \right)^{{t \over {{T_{1/2}}}}}}

⇒\Rightarrow t = 2T 1/2 = 2 ×\times 10 minutes = 20 minutes

Q29
The ratio of kinetic energy to the total energy of an electron in a Bohr orbit of the hydrogen atom, is
A 1 : 1
B 1 : -1
C 2 : -1
D 1 : -2
Correct Answer
Option B
Solution

In a Bohr orbit of the hydrogen atom, Kinetic energy = – (Total energy) So, Kinetic energy : Total energy = 1 : –1

Q30
Radioactive material 'A' has decay constant '8 λ\lambda ' and material 'B' has decay constant 'λ\lambda '. Initially they have same number of nuclei. After what time, the ratio of number of nuclei of material 'B' to that 'A' will be e ?
A 17λ{1 \over {7\lambda }}
B 18λ{1 \over {8\lambda }}
C 19λ{1 \over {9\lambda }}
D 1λ{1 \over {\lambda }}
Correct Answer
Option A
Solution

Ratio of number of nuclei

N1N2=e−8λte−λt{{{N_1}} \over {{N_2}}} = {{{e^{ - 8\lambda t}}} \over {{e^{ - \lambda t}}}}

=

e−7λt{{e^{ - 7\lambda t}}}

According to question,

e−7λt{{e^{ - 7\lambda t}}}

=

1e{1 \over e}

⇒\Rightarrow t =

17λ{1 \over {7\lambda }}
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