Nuclei Test

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Nuclei Test - 24

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Question 1
1 / -0

The binding energy per nucleon of Uranium in $$MeV$$ is
(Atomic mass of uranium $$m_{a}=\ 238.0508 amu;$$
Mass of hydrogen atom is$$M_{H}=\ 1.0078 amu $$;
Mass of neutron $$m_{N}=$$ $$1.0087 amu$$; atomic number of Uranium $$Z $$$$=$$ 92; Mass Number of Uranium $$A$$ $$=$$ 238)

A
$$7.580$$

B
$$6.216$$

C
$$5.162$$

D
$$3.146$$

Solution

No. of protons $$=$$ $$92$$

No. of neutrons $$= 238 - 92 = 146$$

$$\Delta m = [92\times 1.0078 + 146\times 1.0087 - 238.0508] amu$$

$$= 1.937 amu$$

$$BE = 1.937\times 931.478 MeV$$

$$= 1804.27 MeV$$

$$BE/per\ nucleon = \dfrac{1804.27}{238}MeV$$
$$= 7.58MeV$$
Question 2
1 / -0

The energy required to split $$^{16}_{8}O$$ nucleus into four $$\alpha $$ particles is $$\underline{\hspace{0.5in}}$$.
(The mass of an $$\alpha $$ particle is $$4.0026.03 amu$$ and that of oxygen is $$15.994915 amu$$)

A
$$11.6 MeV$$

B
$$13.7 MeV$$

C
$$14.4 MeV$$

D
$$15.8 MeV$$

Solution

$$^{16}_8O \rightarrow 4 ^4_2{He}$$
$$\Delta m = [4\times 4.002603 - 15.994915] \mu$$
$$= 0.015497 \mu$$
$$Q = \Delta m C^2$$
$$= 0.015497\times 931.478\ MeV$$
$$= 14.44 \ MeV$$
Question 3
1 / -0

The kinetic energy of the $$\alpha $$ - particle emitted in the decay $$^{238}_{94}Pu \rightarrow ^{234}_{92}U + $$ $$^4_2He$$.
(The atomic masses of $$^{238}Pu, ^{234}U$$ and $$\alpha$$ particle are 238.04955u, 234.04095 u and 4.002603u respectively. Neglect any recoil of the nucleus)

A
4.392 MeV

B
6.259 MeV

C
5.592 MeV

D
4.952 MeV

Solution

Assuming that energy librated by mass defect is converted into KE of $$\alpha -$$particle.
$$\Delta m = (238.04955 - 234.04095 - 4.002603) \mu$$
$$= 0.005997 \mu$$
$$KE = E = 0.005997\times 931.478\ MeV$$
$$KE = {5.586} \ MeV$$
Question 4
1 / -0

The energy released in the following process is :
$$A + B $$ $$\rightarrow $$$$ C + D +Q$$
(mass of $$A$$ is $$1.002$$ amu ; mass of $$B$$ is $$1.004$$ amu ;mass of $$C$$ is $$1.001$$ amu ; mass of $$D$$ is $$1.003$$ amu)

A
1.254 MeV

B
0.931 MeV

C
0.465 MeV

D
1.862 MeV

Solution

$$\Delta m = [1.002 + 1.004 - 1.001 - 1.003]$$
$$ = 0.002 \mu$$

$$E = Q = 0.002\times 931.478 MeV$$
$$= 1.862 MeV$$
Question 5
1 / -0

Mass of proton $$= 1.00760 amu$$, mass of neutron $$= 1.00899 amu$$, mass of deuterium nucleus $$=2.0147 amu$$. Then binding energy is :

A
$$0.00189 MeV$$

B
$$1.76 amu$$

C
$$1.76 MeV$$

D
$$10^{15} joules$$

Solution

$$^2_1H \rightarrow 1 proton + 1 neutron$$

$$\Delta m = [1.0076 + 1.00899 + 2.0147]$$
$$= 0.00189 \mu$$

$$E = 0.00189\times 931.478 MeV$$
$$= 1.76 MeV$$
Question 6
1 / -0

In nuclei with mass number greater than $$20$$, the average binding energy is:

A
$$8MeV$$

B
$$0.8MeV$$

C
$$80MeV$$

D
$$0.08MeV$$

Solution

The binding energy per nucleon Vs number of nucleon curve suggests that average binding is almost $$8\ MeV$$ for the nuclei having mass number greater than $$20$$.
Question 7
1 / -0

$$_{3}Li^{7}+_{1}H^{2}\rightarrow _{4}Be^{8}+_{o}n^{1}+Q$$
Mass of $$_{3}Li^{7}=$$ $$7.01823amu$$
Mass of $$_{1}H^{2}=$$ $$2.01474amu$$
Mass of $$_{4}Be^{8}=$$$$8.00785 amu$$
Mass of $$_{o}n^{1}=$$ $$1.00899 amu$$
Then, the value of Q is

A
$$5\ MeV$$

B
$$10\ MeV$$

C
$$15\ MeV$$

D
$$0$$

Solution

Mass of reactants is: $$7.01824 + 2.01474 = 9.03297\ amu$$
Mass of products is: $$8.00785 + 1.00899 = 9.01684\ amu$$
The difference in mass $$= 0.01613\ amu$$. (This mass has been converted to energy).
$$1 amu = 1.67 \times 10^{-27} kg$$
Hence, mass difference is $$= 0.01613 \times 1.67 \times 10^{-27} kg = 2.69371 \times 10^{-29} kg$$
Using Einstein's relation:
$$E = mc^{2}$$
$$E = 2.69371 \times 10^{-21} \times (3 \times 10^{8})^{2} = 2.42 \times 10^{-12} J$$
$$E = 15\ MeV$$
Question 8
1 / -0

The mass defect and binding energy of $$^{12}_{6}$$C nucleus is
(Mass of $$^{12}_{6}$$ C $$=$$12.000000 amu; $$m_{p}=$$1.007825 amu and $$m_{n}=$$1.008665 amu)

A
$$0.06522$$ amu; $$72.1$$ MeV

B
$$0.09894$$ amu; $$92.1$$ MeV

C
$$0.05315$$ amu; $$102.2$$ MeV

D
$$0.05315$$ amu; $$82.2$$ MeV

Solution

C $$=$$ 6 neutrons + 6 protons
$$\Delta m = (6\times 1.007825 + 6\times 1.008665 - 12)$$
$$= 0.09894 \ amu$$
$$E = (0.09894\times 931.478 \dfrac{MeV}{C^2})C^2$$
$$= 92.1 MeV$$
Question 9
1 / -0

The binding energy per nucleon of $$^{40}_{20}Ca$$ is $$\underline{\hspace{0.5in}}$$
($$^{40}_{20}Ca =$$39.962589 amu, $$m_{p} =$$1.007825 amu; $$m_{n}=$$$$1.008665 amu$$ and $$1amu$$ is equivalent to $$931.5 MeV$$)

A
$$6.55 MeV$$

B
$$7.55 MeV$$

C
$$8.55 MeV$$

D
$$9.55 MeV$$

Solution

$$\Delta m = [20\times 1.007825 + 20\times 1.008665 - 39.962589] \mu$$
$$= 0.367211 \mu$$

$$BE = 0.367211\times 931.478 MeV$$
$$= 342.045 MeV$$

BE per nucleon $$= \dfrac{BE}{40} = 8.55\ MeV$$
Question 10
1 / -0

The binding energy per nucleon of $$^{35}_{17}Cl$$ nucleus is
($$^{35}_{17}Cl =$$34.98000 amu, $$m_{P}=$$1.007825 amu,$$m_{n} =$$1.008665 amu and 1 amu is equivalent to 931MeV)

A
$$4.6 MeV$$

B
$$5.8 MeV$$

C
$$6.5 MeV$$

D
$$8.2 MeV$$

Solution

In $$Cl$$ there are $$17$$ protons and $$(35 - 17)$$ $$=$$ 18 neutrons.

$$\Delta m = [17\times 1.007825 + 18\times 1.008665 - 34.98] amu$$

$$= 0.308995$$

$$BE = 0.308995\times 931.478 MeV$$

$$= 287.82 MeV$$

$$\dfrac{BE}{nucleon} = \dfrac{287.82}{35} = 8.22\ MeV$$

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Nuclei Test - 24

Result

Nuclei Test - 24

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SHARING IS CARING

Question 1

$$7.580$$

$$6.216$$

$$5.162$$

$$3.146$$

Solution

Question 2

$$11.6 MeV$$

$$13.7 MeV$$

$$14.4 MeV$$

$$15.8 MeV$$

Solution

Question 3

4.392 MeV

6.259 MeV

5.592 MeV

4.952 MeV

Solution

Question 4

1.254 MeV

0.931 MeV

0.465 MeV

1.862 MeV

Solution

Question 5

$$0.00189 MeV$$

$$1.76 amu$$

$$1.76 MeV$$

$$10^{15} joules$$

Solution

Question 6

$$8MeV$$

$$0.8MeV$$

$$80MeV$$

$$0.08MeV$$

Solution

Question 7

$$5\ MeV$$

$$10\ MeV$$

$$15\ MeV$$

$$0$$

Solution

Question 8

$$0.06522$$ amu; $$72.1$$ MeV

$$0.09894$$ amu; $$92.1$$ MeV

$$0.05315$$ amu; $$102.2$$ MeV

$$0.05315$$ amu; $$82.2$$ MeV

Solution

Question 9

$$6.55 MeV$$

$$7.55 MeV$$

$$8.55 MeV$$

$$9.55 MeV$$

Solution

Question 10

$$4.6 MeV$$

$$5.8 MeV$$

$$6.5 MeV$$

$$8.2 MeV$$

Solution

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