Nuclei Test

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Nuclei Test - 30

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Question 1
1 / -0

Two deutrons fuse to form a helium nucleus and energy is released, because the mass of helium nucleus is

A
equal to that of two deutrons

B
less than that of two deutrons

C
more than that of two deutrons

D
all the above

Solution

Mass of deuteron nucleus is 2.0141 amu and masss of helium nucleus is 4.0026 amu. When two deuterons fuse to form a helium nucleus some of the mass will be lost and the mass defect is
$$\Delta m = (2 \times 2.0141) - 4.0026$$
$$\Delta m = 4.0282 - 4.0026$$
$$\Delta m = 0.0256 amu$$
This mass is converted into energy and is released.
Hence, two deuterons fuse to form a helium nucleus and energy is released, because the mass of helium nucleus is less than that of two deuterons
Question 2
1 / -0

In the fusion process there are

A
isotopes of hydrogen

B
isotopes of helium

C
isotopes of carbon

D
isotopes

Solution

The lighter nuclei have smaller binding energies per particle than heavier one. The lighter nuclei combine to form the heavier nuclei i.e. fusion reaction.
Amongst isotopes of hydrogen, helium and carbon, hydrogen isotopes(deuterium has 1 prioton, 1 neutron and tritium has 1,proton and 2 neutrons) are lighter since their atomic mass is one.
Question 3
1 / -0

The mass defect for helium nucleus is $$0.0304$$ $$a.m.u$$. The binding energy per nucleon of helium nucleus is ________

A
$$28.3$$ $$MeV$$

B
$$7.075$$ $$MeV$$

C
$$9.31$$ $$MeV$$

D
$$200$$ $$MeV$$

Solution

$$1\ amu = 1.67 \times 10^{-27} kg$$
Using this and Einsteins relation, we get the energy equivalent to the rest mass as:
$$E = mc^{2} = (1.67 \times 10^{-27} \times 0.0304) \times (3 \times 10^{8})^{2} =4.569 \times 10^{-12}J$$

Converting this energy in electron volts.
$$E = \dfrac{4.569 \times 10^{-12}}{1.6 \times 10^{-19}} = 28MeV$$

This is per nucleon of Helium. One Helium nucleus has four nucleons.

Hence, binding energy per nucleon is $$7\ MeV$$.
Question 4
1 / -0

The necessary condition for nuclear fusion is

A
high temperature and high pressure

B
low temperature and low pressure

C
high temperature and low pressure

D
low temperature and high pressure

Solution

High temperature - The high temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons.
High pressure - Pressure squeezes the hydrogen atoms together. They must be within $$1\times10^{-15}$$ meters of each other to fuse.
Question 5
1 / -0

The fusion process is possible at high temperature because at high temperatures

A
the nucleus disintegrates

B
molecules disintegrate

C
atoms become ionised

D
the nuclei get sufficient energy so as to overcome the Coulomb repulsive force

Solution

High temperature - The high temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons.
Question 6
1 / -0

For the fast chain reaction, the size of $$U^{235}$$ block, as compared to its critical size, must be

A
greater

B
smaller

C
same

D
anything

Solution

The critical size must at least include enough fissionable material to reach critical mass. If the size of the reactor core is less than a certain minimum, fission neutrons escape through its surface and the chain reaction is not sustained.
Question 7
1 / -0

The critical mass of fissionable material is

A
$$75$$ $$kg$$

B
$$1$$ $$kg$$

C
$$20$$ $$kg$$

D
$$10$$ $$kg$$

Solution

A critical mass is the smallest amount of fissile material needed for a sustained nuclear chain reaction.
In the case of Plutonium-239, the critical mass is about 10 kg.
Question 8
1 / -0

The value of A in the following reaction is
$$_{4}\textrm{Be}^{9}$$ + $$_{2}\textrm{He}^{4}$$ = $$_{6}\textrm{C}^{A}$$ + $$_{0}\textrm{n}^{1}$$

A
$$14$$

B
$$10$$

C
$$12$$

D
$$16$$

Solution

The reaction producing neutrons by bombarding beryllium with alpha particles is given as
$$_{4}^{9}\textrm{Be} + _{2}^{4}\textrm{He} \rightarrow _{6}^{A}\textrm{C} + _{0}^{1}\textrm{n}$$
Since, a neutron has atomic number of zero (it has no positive charge) but has a mass number of one (the same mass as the proton). Hence, the total mass (that is, the total number of nuclear particles or nucleons = 13) entering into the reaction must be the same as nucleons leaving the reaction.
Therefore $$A = 12$$.
$$_{4}^{9}\textrm{Be} + _{2}^{4}\textrm{He} \rightarrow _{6}^{12}\textrm{C} + _{0}^{1}\textrm{n}$$
Thus atomic mass is conserved in the reaction.
Question 9
1 / -0

When the number of nucleons in the nucleus increased then the binding energy per nucleon

A
decreases continuously with A

B
increases continuously with A

C
remains constant with A

D
first increases with A and then decreases

Solution

Binding energy per nucleon Vs number of nucleons curve is shown above which suggests that binding energy per nucleon increases initially (upto Fe) and then decreases.
Question 10
1 / -0

The curve between binding energy per nucleon $$(E)$$ and mass number $$(A)$$ is:

A

B

C

D

Solution

We know that the binding energy per nucleon (E) increases upto $$A=56$$ then it decreases as A increases. So $$^{56}Fe $$ is the most stable element. Thus, option C will represent correct graph.