Atoms or Nuclear Fission and Fusion MCQ Questions & Answers in Modern Physics | Physics
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61.
Fusion reaction takes place at high temperature because
A
atoms get ionised at high temperature
B
kinetic energy is high enough to overcome the coulomb repulsion between nuclei
C
molecules break up at high temperature
D
nuclei break up at high temperature
Answer :
kinetic energy is high enough to overcome the coulomb repulsion between nuclei
Fusion reaction takes place at high temperature because kinetic energy is high enough to overcome the couloumb repulsion between nuclei.
62.
The nuclear radius of $$_8{O^{16}}$$ is $$3 \times {10^{ - 15}}m.$$ If an atomic mass unit is $$1.67 \times {10^{ - 27}}kg,$$ then the nuclear density is approximately
63.
What is the radius of iodine atom ?
(atomic no. 53, mass no. 126)
A
$$2.5 \times {10^{ - 11}}m$$
B
$$2.5 \times {10^{ - 9}}m$$
C
$$7 \times {10^{ - 9}}m$$
D
$$7 \times {10^{ - 6}}m$$
Answer :
$$2.5 \times {10^{ - 11}}m$$
Electronic configuration of iodine $$\left( {53} \right)$$ is $$= 2, 8, 18, 18, 7$$
∴ Principal quantum number $$n = 5$$
Radius of $$n$$th orbit is given by
$${r_n} = {r_0}\left( {\frac{{{n^2}}}{Z}} \right)\,\,\left( {Z = {\text{atomic number}}} \right)$$
where $${r_0} = 0.53\,\mathop {\text{A}}\limits^ \circ = 0.53 \times {10^{ - 10}}m$$
$$\therefore {r_n} = \left( {0.53 \times {{10}^{ - 10}}} \right) \times \frac{{{5^2}}}{{53}} = 2.5 \times {10^{ - 11}}m$$
64.
The nuclear fusion reaction $${2_1}{H^2}{ \to _2}H{e^4} + {\text{Energy}},$$ is proposed to be used for the production of industrial power. Assuming the efficiency of process for production of power is $$20\% ,$$ find the mass of the deuterium required approximately for a duration of 1 year. Given mass of $$_1{H^2}\,{\text{nucleus}} = 2.0141\,a.m.u$$ and mass of $$_2H{e^4}\,{\text{nuclei}} = 4.0026\,a.m.u$$ and $$1\,a.m.u = 31\,MeV$$
A
$$165\,kg$$
B
$$138\,kg$$
C
$$180\,kg$$
D
$$60\,kg$$
Answer :
$$138\,kg$$
Mass defect $$\Delta m = 2 \times 2.014 - 4.0026 = 0.0256\,a.m.u$$
Energy released when two $$_1{H^2}$$ nuclei fuse $$ = 0.0256 \times 931 = 23.8\,MeV$$
Total energy required to be produced by nuclear reaction in 1 year
$$ = 2500 \times {10^6} \times 3.15 \times {10^7} = 7.88 \times {10^{16}}J$$
No. of nuclei of $$_1{H^2}$$ required
$$ = \frac{{7.88 \times {{10}^{16}}J}}{{23.8 \times 1.6 \times {{10}^{ - 13}}}} \times 2 = 4.14 \times {10^{28}}$$
Mass of Deuterium required
$$ = \frac{{4.14 \times {{10}^{28}}}}{{6.02 \times {{10}^{23}}}} \times 2 \times {10^{ - 3}}kg = 138\,kg$$
65.
Imagine that a reactor converts all given mass into energy and that it operates at a power level of $${10^9}watt.$$ The mass of the fuel consumed per hour in the reactor will be : (velocity of light, $$c$$ is $$3 \times {10^8}m/s$$ )
A
$$0.96\,gm$$
B
$$0.8\,gm$$
C
$$4 \times {10^{ - 2}}gm$$
D
$$6.6 \times {10^{ - 5}}gm$$
Answer :
$$4 \times {10^{ - 2}}gm$$
Power level of reactor, $$P = \frac{E}{{\Delta t}} = \frac{{\Delta m{c^2}}}{{\Delta t}}$$
mass of the fuel consumed per hour in the reactor,
$$\frac{{\Delta m}}{{\Delta t}} = \frac{P}{{{c^2}}} = \frac{{{{10}^9}}}{{{{\left( {3 \times {{10}^8}} \right)}^2}}}kg/\sec = \frac{{3600 \times {{10}^9}}}{{9 \times {{10}^{16}}}} \times {10^3}\frac{{gm}}{{hr}}$$