Nuclei Test

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Nuclei Test - 23

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Weekly Quiz Competition

Question 1
1 / -0

The overall process of carbon nitrogen fusion cycle results in the fusion of 4 protons to yield helium nucleus and :

A
positron

B
two electrons

C
two positrons

D
an electron

Solution

Overall reaction of carbon nitrogen cycle
$$4 _1^1H \ \rightarrow \ _2^4He + \ 2(_1^0e) + \ Q$$
(Proton) (Helium) (Positron) (energy)
So, 2 positrons are released.
Question 2
1 / -0

In Carbon-Nitrogen fusion cycle , protons are fused to form a helium nucleus, positrons and release some energy.The number of protons fused and the number of positrons released in this process respectively are

A
4,4

B
4,2

C
2,4

D
4,6

Solution

Overall reaction of Carbon - Nitrogen cycle is
$$4 _1^1H \rightarrow _2^4He + 2(_1^0e) + Q$$
So, 4 protons are fused to give 2 positrons.
Question 3
1 / -0

Energy in the sun is due to

A
Fossil fuels

B
Radioactivity

C
Fission

D
Fusion

Solution

Nuclear fusion takes place in the sun and stars.
So, energy in the sun is due to fusion.
Question 4
1 / -0

$$1$$ kg of iron (specific heat $$120$$ Cal kg$$^{-1}$$C$$^{-1}$$) is heated by $$1000$$$$^{0}$$C. The increase in its mass is :

A
Zero

B
$$5.6 \times 10$$$$^{-8}$$ Kg

C
$$5.6 \times 10$$$$^{-16}$$ Kg

D
$$5.6 \times 10$$$$^{-12 }$$g

Solution

Energy given to it $$ = mS\Delta T$$
$$ = (1 kg)(120 \dfrac{cal}{Kg C})(1000^{0}C)$$
$$ = 12 \times 10^{4} Cal$$
$$ = 12 \times 10^{4} \times 4.18 J$$
$$ = 50 \times 10^{4} J$$

The energy given to it is used to increase its temperature by $$1000^{0} C$$, therefore mass increment is ZERO.
Question 5
1 / -0

The mass defect for the nucleus of Helium is $$0.0303 \ amu$$. The binding energy per nucleon in $$MeV$$ is

A
28

B
7

C
4

D
1

Solution

$$ B.E = (931.478 \times .0303 \dfrac{MeV}{c^{2}}) c^{2}$$
$$ = 28.22 MeV$$

$$ B.E\ per\ nucleon = \dfrac{BE}{A} = \dfrac{28.22}{4} = 7.05 MeV$$
Question 6
1 / -0

The mass defect and binding energy per nucleon of an alpha particle are :
(Mp $$=$$ 1.00734$$u$$, Mn $$=$$ 1.00874$$u$$, M $$=$$ 4.0015$$u$$)

A
$$0.0303u, 28.3 MeV$$

B
$$0.1303 u, 28.3 MeV$$

C
$$0.0303u, 7.07 MeV$$

D
$$0.0306u, 7.14 MeV$$

Solution

$$^4_2He$$

$$\Delta m = [2\times 1.00734 + 2\times 1.00874 - 4.0015]$$$$= 0.03066 u$$

$$B.E = 0.03066\times 931.478 MeV$$$$= 28.56 MeV$$

$$ B.E \ per\ nucleon = \dfrac{28.56}{4}$$$$= {7.14 MeV}$$
Question 7
1 / -0

The binding energy of an imaginary iron $$^{56}_{36}Fe$$ is $$\underline{\hspace{0.5in}}$$
(Given atomic mass of Fe is 55.9349 amu and that of hydrogen is 1.00783 amu. Mass of neutron is 1.00876 amu)

A
$$3.49 MeV$$

B
$$4.31MeV$$

C
$$6.49 MeV$$

D
$$931.49 MeV$$

Solution

The number of proton of Fe nucleus $$p=36$$ and number of neutron, $$n=56-36=20$$
The combined mass of nucleus $$=$$ total mass of proton $$+$$ total mass of neutron
$$=(36\times 1.00783) +(20\times 1.00876)\\ =56.4571 $$ amu

Mass defect $$\Delta m=56.457-55.9349=0.522$$ amu
Binding energy, $$BE={\Delta m c^2}$$
So, $$BE=0.522 c^2$$ amu $$=0.522\times 931.49 MeV=486.24$$ MeV where, $$1 amu = 931.49 MeV/c^2$$
Question 8
1 / -0

The masses of $$\alpha $$ -particle, proton and neutron are $$4.00150$$ amu, $$1.00728$$ amu and $$1.00867$$ amu respectively. Binding Energy per nucleon of $$\alpha$$ -particle is :

A
$$283 J$$

B
$$931.5 MeV$$

C
$$28.3 MeV$$

D
$$7.08 MeV$$

Solution

$$2$$ proton, $$2$$ neutron

$$\Delta m = [2\times 1.00728 + 2\times 1.00867 - 4.00150] amu$$
$$= 0.0304 amu$$

$$BE = 0.0304\times 931.478 MeV$$
$$= 28.32 MeV$$

Average $$BE = \dfrac{28.32}{4} = 7.08\ MeV$$
Question 9
1 / -0

The mass defect in $$_{2}He^{3}$$ is $$\underline{\hspace{0.5in}}$$ , if m$$_{p} =$$1.00727 amu, $$m_{n} =$$1.008665 amu, mass of $$_{2}He^{3} =$$3.01664 amu

A
0.00657 amu

B
0.00621 amu

C
0.0657 amu

D
0.6877 amu

Solution

The difference between atomic weight and mass number i.e. mass of elementary particle in the nucleus is known as mass defect. Since, $$_{3}^{2}\textrm{He}$$ has two protons(atomic number = proton) and one neutron(mass number = proton + neutron).
Hence, Mass defect is
$$\Delta m = [2(m_p) + 1(m_n)] - M_{He}$$
where, variables have their usual meanings.
$$\Delta m = [2(1.00727) + 1(1.008665)] - 3.01664$$
$$\Delta m = [2.01454 + 1.008665] - 3.01664$$
$$\Delta m = 3.023205 - 3.01664$$
$$\Delta m = 0.006565 amu$$
$$\Delta m \approx 0.00657 amu$$
Question 10
1 / -0

The binding energy of an $$\alpha $$ -particle is
(Given that mass of proton$$=$$$$1.0073$$ amu, mass of neutron$$=$$ $$1.0087$$ amu, and mass of $$\alpha $$-particle$$=$$$$4.0015$$ amu)

A
$$28.4 MeV$$

B
$$35.6 MeV$$

C
$$40.3 MeV$$

D
$$44.7 MeV$$

Solution

$$\Delta m = (2\times 1.0073 + 2\times 1.0087 - 4.0015)$$
$$= 0.00305 \mu$$
$$E = 0.0305\times 931.478 MeV$$
$$= 28.41\ MeV$$

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Nuclei Test - 23

Result

Nuclei Test - 23

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Question 1

positron

two electrons

two positrons

an electron

Solution

Question 2

4,4

4,2

2,4

4,6

Solution

Question 3

Fossil fuels

Radioactivity

Fission

Fusion

Solution

Question 4

Zero

$$5.6 \times 10$$$$^{-8}$$ Kg

$$5.6 \times 10$$$$^{-16}$$ Kg

$$5.6 \times 10$$$$^{-12 }$$g

Solution

Question 5

28

7

4

1

Solution

Question 6

$$0.0303u, 28.3 MeV$$

$$0.1303 u, 28.3 MeV$$

$$0.0303u, 7.07 MeV$$

$$0.0306u, 7.14 MeV$$

Solution

Question 7

$$3.49 MeV$$

$$4.31MeV$$

$$6.49 MeV$$

$$931.49 MeV$$

Solution

Question 8

$$283 J$$

$$931.5 MeV$$

$$28.3 MeV$$

$$7.08 MeV$$

Solution

Question 9

0.00657 amu

0.00621 amu

0.0657 amu

0.6877 amu

Solution

Question 10

$$28.4 MeV$$

$$35.6 MeV$$

$$40.3 MeV$$

$$44.7 MeV$$

Solution

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