Nuclei Test

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Nuclei Test - 57

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Weekly Quiz Competition

Question 1
1 / -0

Consider the following reaction:
$${ _{ }^{ 1 }{ H } }_{ 2 }+{ _{ }^{ 1 }{ H } }_{ 2 }\rightarrow { _{ 1 }^{ }{ He } }^{ 4 }+Q$$
If $$m({ _{ 1 }^{ }{ H } }^{ 2 })=2.0141u$$; $$m({ _{ 2 }^{ }{ He } }^{ 4 })=4.0024u$$, the energy $$Q$$ released (in MeV) in this fusion reacion is

A
12

B
6

C
24

D
48

Solution

$${ _{ 1 }^{ }{ H } }^{ 2 }+{ _{ 1 }^{ }{ H } }^{ 2 }\rightarrow { _{ 1 }^{ }{ He } }^{ 4 }+Q\quad $$
$$\Delta m=m\left( { _{ 2 }^{ }{ He } }^{ 4 } \right) -2m\left( { _{ 1 }^{ }{ H } }^{ 2 } \right) =-0.0258u$$
$$Q={ c }^{ 2 }\Delta m$$
$$=\left( 0.0258 \right) \left( 931.5 \right) MeV$$
$$\approx 24MeV$$
Question 2
1 / -0

In the nuclear reaction
$${ _{ 1 }^{ }{ H } }^{ 2 }+{ _{ 1 }^{ }{ H } }^{ 2 }\rightarrow { _{ 2 }^{ }{ He } }^{ 3 }+{ _{ 0 }^{ }{ n } }^{ 1 }$$
if the mass of the deuterium atom $$=2.014741a.m.u$$, mass of $${ _{ 2 }^{ }{ He } }^{ 3 }=3.016977a.m.u$$ and mass of neutron $$=1.008987a.m.u$$, then the $$Q$$ value of the reaction is nearly

A
$$0.00352MeV$$

B
$$3.27MeV$$

C
$$0.82MeV$$

D
$$2.45MeV$$

Solution

$$Q=-\Delta mc^2$$
$$Q=\left( \sum { { M }_{ r } } -\sum { { M }_{ p } } \right) { c }^{ 2 }$$
$$\sum { { B }_{ r } } =2\times 2.014741a.m.u=4.0294892a.m.u$$
$$\sum { { B }_{ p } } =(3.016977+1.008987)a.m.u=4.025964a.m.u$$
$$\sum { { B }_{ r } } -\sum { { B }_{ p } } =0.003518a.m.u\quad $$

The decrease in mass appears as equivalent energy
$$Q=0.003518\times 931MeV=3.27MeV\quad $$
Question 3
1 / -0

Binding energy per nucleon for $${C}^{12}$$ is $$7.68MeV$$ and for $${C}^{13}$$ is $$7.74MeV$$. The energy required to remove a neutron from $${C}^{13}$$ is

A
$$5.49MeV$$

B
$$8.46MeV$$

C
$49.45MeV$$

D
$$15.49MeV$$

Solution

$$C^{13}\longrightarrow C^{12}+n$$
Binding energy of $$C^{12}=12\times7.68MeV=92.16MeV$$
Binding energy of $$C^{13}=13\times7.74MeV=100.62MeV$$

Energy required to remove a neutron $$=B_{C^{13}}-B_{C^{12}}$$
$$E=100.62MeV-92.16MeV$$
$$E=8.46MeV$$
Question 4
1 / -0

Binding energy per nucleon of $${ _{ 1 }^{ }{ H } }^{ 2 }$$ and $${ _{ 2 }^{ }{ He } }^{ 4 }$$ are $$1.1MeV$$ and $$7.0MeV$$, respectively. Energy released in the process $${ _{ 1 }^{ }{ H } }^{ 2 }+{ _{ 1 }^{ }{ H } }^{ 2 }\rightarrow{ _{ 2 }^{ }{ He } }^{ 4 }$$ is

A
$$20.8MeV$$

B
$$16.6MeV$$

C
$$25.2MeV$$

D
$$23.6MeV$$

Solution

Given
$$BE_{He}=4\times7.0MeV=28.0MeV$$
$$BE_H=2\times1.1Mev=2.2MeV$$

Energy released in the reaction is equal to the total binding energy of reactants minus the total binding energy of products.
Hence, Energy released would be
$$\Delta E=BE_{He}-2\times BE_{H}$$
$$\Delta E=(28.0-2\times 2.2)MeV$$
$$\Delta E=23.6MeV$$
Question 5
1 / -0

When a nucleus in an atom undergoes radioactive decay, the electronic energy levels of the atom

A
do not change for any type of radioactivity

B
change for $$\alpha$$ and $$\beta$$ radioactivity but not for $$\gamma$$- radioactivity

C
change for $$\alpha$$- radioactivity but not for others

D
change for $$\beta$$- radioactivity but not for others

Solution

$$\beta$$ particles carries one unit of negative charge, and $$\alpha$$ particle carries $$2$$ units of positive charge, and Y-particle carries no charge. So the electronic energy level of the atom changes in emission of $$\alpha$$ and $$\beta$$ particle, but not in $$Y$$ decay.
Question 6
1 / -0

Tritium is an isotope of hydrogen whose nucleus Triton contains $$2$$ neutrons and $$1$$ proton. Free neutrons decay into $$p+\vec { e } +\vec { v } $$. If one of the neutrons in Triton decays, it would transform into $${ _{ 2 }^{ }{ He } }^{ 3 }$$. This does not happen. This is because

A
Triton energy is less than that of $${He}^{3}$$ nucleus

B
the electron created in the beta decay process cannot remain in the nucleus

C
both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a $${He}^{3}$$ nucleus

D
because free neutrons decay due to external pertubations which is absent in triton nucleus

Solution

Triton $$\left( { _{ 1 }^{ }{ H } }^{ 3 } \right) $$ has 1 proton and 2 neutrons. If a neutron decays as $$\rightarrow p+\vec { e } +\vec { v } $$, then nucleus will have 2 proton and 1 neutron, ie. triton atom converts in $${ _{ 2 }^{ }{ He } }^{ 3 }$$ (2 proton and 1 neutron)
Binding energy of $$\left( { _{ 1 }^{ }{ H } }^{ 3 } \right) $$ is much smaller than $${ _{ 2 }^{ }{ He } }^{ 3 }$$ so transformation is not possible energetically
Question 7
1 / -0

Heavy stable nucleus have more neutrons than protons. This is because of the fact that

A
neutrons are heavier than protons

B
electrostatic force between protons are repulsive

C
neutrons decay into protons through beta decay

D
nuclear force between neutrons are weaker than that between protons

Solution

Electrostatic force between proton-proton is repulsive which causes the un-stability of nucleus. So, neutrons ar larger than proton
Question 8
1 / -0

The particle used in nuclear fission for bombardment is:

A
alpha particle

B
proton

C
beta particle

D
neutron.

Solution

Nuclear fission occurs when an atom splits into two or more smaller atoms, most often the as the result of neutron bombardment.
Question 9
1 / -0

Nuclear binding energy is equivalent to

A
Mass of proton

B
Mass of neutron

C
Mass of nucleus

D
Mass detect of nucleus

Solution

To find the binding energy, add the masses of the individual protons, neutrons, and electrons, subtract the mass of the atom, and convert that mass difference to energy.
Question 10
1 / -0

The accepted unit of atomic and molecular mass is:

A
kilogram

B
gram

C
pound

D
atomic mass unit

Solution

Accepted as unit of atomic and molecular mass is atomic mass unit (u).

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Nuclei Test - 57

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Nuclei Test - 57

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SHARING IS CARING

Question 1

12

6

24

48

Solution

Question 2

$$0.00352MeV$$

$$3.27MeV$$

$$0.82MeV$$

$$2.45MeV$$

Solution

Question 3

$$5.49MeV$$

$$8.46MeV$$

$49.45MeV$$

$$15.49MeV$$

Solution

Question 4

$$20.8MeV$$

$$16.6MeV$$

$$25.2MeV$$

$$23.6MeV$$

Solution

Question 5

do not change for any type of radioactivity

change for $$\alpha$$ and $$\beta$$ radioactivity but not for $$\gamma$$- radioactivity

change for $$\alpha$$- radioactivity but not for others

change for $$\beta$$- radioactivity but not for others

Solution

Question 6

Triton energy is less than that of $${He}^{3}$$ nucleus

the electron created in the beta decay process cannot remain in the nucleus

both the neutrons in triton have to decay simultaneously resulting in a nucleus with 3 protons, which is not a $${He}^{3}$$ nucleus

because free neutrons decay due to external pertubations which is absent in triton nucleus

Solution

Question 7

neutrons are heavier than protons

electrostatic force between protons are repulsive

neutrons decay into protons through beta decay

nuclear force between neutrons are weaker than that between protons

Solution

Question 8

alpha particle

proton

beta particle

neutron.

Solution

Question 9

Mass of proton

Mass of neutron

Mass of nucleus

Mass detect of nucleus

Solution

Question 10

kilogram

gram

pound

atomic mass unit

Solution

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