Nuclei Test

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Nuclei Test - 61

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Question 1
1 / -0

The rest mass of a deuteron is equivalent to an energy of $$1876$$ MeV, that of a proton to $$939$$ MeV, and that of a neutron to $$940$$ MeV. A deuteron may disintegrate to a proton and a neutron if it

A
emits an X-ray photon of energy $$2$$ MeV

B
captures an X-ray photon of energy $$2$$ MeV

C
emits an X-ray photon of energy $$ 3$$ MeV

D
captures an X-ray photon of energy $$3$$ MeV

Solution

DEUTRON
$$_1 H^2 = n+p$$
Total energy of deutron $$= 1476\ MeV$$
Energy of neutron $$+$$ proton $$= 940 +939 = 1879\ MeV$$
Energy required to disintegrate Deutron $$= 3\ MeV$$
Question 2
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A helium atom, a hydrogen atom and a neutron have masses of $$4.003 u$$, $$1.008 u$$ and $$1.009 u$$ (unified atomic mass units), respectively. Assuming that hydrogen atoms and neutrons can fuse to form helium, what is the binding energy of a helium nucleus?

A
$$2.01 u$$

B
$$3.031 u$$

C
$$1.017 u$$

D
$$0.031 u$$

Solution

$$ 2 \:_1H^1 + 2n \rightarrow _2 He ^4$$
Binding energy $$=$$ Mass defect
$$=-m_{He} + 2m_n + 2m_H$$
$$ =2(1.008) + 2(1.009) -4.003 $$
$$=0.031 u$$
Question 3
1 / -0

Assuming that about 20 MeV of energy is released per fusion reaction $$_1H^2+_1H^2\rightarrow _2He^4+E+$$ other particles, then the mass of $$_1H^2$$ consumed per day in a fusion reactor of power 1 megawatt will approximately be

A
0.001 g

B
1 g

C
10.0 g

D
1000 g

Solution

$$E = \Delta m c^2$$
For one reaction
Mass Defect = $$\Delta m$$
$$ =2(m_{H}) - m_{He} - m_n$$
$$ =2(2.015)-3.017-1.009$$ amu
$$ =0.004 amu$$
$$ 1 amu = 931.5 MeV/c^2$$
Hence
$$E = 0.004 \times 931.5 MeV = 3.724 MeV$$
$$E = 3.726 \times 1.6 \times 10^{-13} J =5.96 \times 10^{-13} J$$
.
Total Requirement = $$ 1 MW = 10^6 J/s$$
Total fusions required = $$ \cfrac{10^6}{5.96 \times 10^{-13} } = 1.67 \times 10^{18}s^{-1}$$
Mass rate =$$ \cfrac{1.67 \times 10^{18}}{N_A} \times 2 \times 2 = 1.1 \times 10^{-5} g/s$$
Total consumption in a day = $$ M \times(24\times 3600) = 0.95 gms$$
Question 4
1 / -0

If mass of $$U^{235}=235.12142amu$$, mass of $$U^{236}=236.1205 amu$$, and mass of neutron $$1.008665 amu$$, then the energy required to remove one neutron from the nucleus of $$U^{236}$$ is nearly about

A
$$75 MeV$$

B
$$6.5 MeV$$

C
$$1 eV$$

D
zero

Solution

The total mass of $$U^{235}$$ and a neutron is $$M_1=235.12142-1.008665=236.130\ amu $$
Increases in mass when one neutron is removed from $$U^{236}=M_1-$$mass of $$U^{236}$$
$$=236.130-236.1205\\ =0.009\ amu \\ =0.009\times 931\ MeV \\ =8.9\ MeV\\ \sim 6.5\ MeV$$
Question 5
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In the nuclear reaction given by $$_2He^4+_7N^{14}\rightarrow _1H^1+X$$ the nucleus X is

A
nitrogen of mass $$16$$

B
nitrogen of mass $$17$$

C
oxygen of mass $$16$$

D
oxygen of mass $$17$$

Solution

$$ _2 He^4 + _7N^{14} \rightarrow _1H^1 + _ZX^A$$

Mass Balance:
$$ 4 + 14 = 1+ A$$
$$A = 17$$

Charge balance:
$$2 + 7 = 1 +Z$$
$$Z =8$$

Element of atomic no. $$8$$ and mass $$17$$ - Oxygen
Question 6
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The binding energies per nucleon of deutreron $$(_1H^2)$$ and helium $$(_2He^4)$$ atoms are $$1.1 MeV$$ and $$7 MeV$$. If two deuteron atoms react to form a single helium atom, then the energy released is

A
$$13.9 MeV$$

B
$$26.9 MeV$$

C
$$23.6 MeV$$

D
$$19.2 MeV$$

Solution

$$ 2 _1 H^2 \rightarrow _2He ^4$$
Energy released $$=$$ Nuclear energy of Product $$-$$ Nuclear energy of Reactant
$$= 7 \times 4 - (1.1 \times 2)\times 2\ MeV$$
$$ = 23.6 \ MeV$$
Question 7
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What would be the energy required to dissociate completely $$1 g$$ of $$Ca^{40}$$ into its constituent particles?
Given: Mass of proton $$=1.007277 amu$$,
Mass of neutron $$=1.00866 amu$$,
Mass of $$Ca-40=39.97545 amu$$
(Take $$1 amu=931 MeV)$$

A
$$4.813\times 10^{24} MeV$$

B
$$4.813\times 10^{24} eV$$

C
$$4.813\times 10^{23} MeV$$

D
none of the above

Solution

$$ Ca = 20p+20n$$

For $$1$$ atom of $$Ca$$:
Mass Defect $$= 20\times 1.007277 + 20\times 1.00866 -39.97545 =40.31-39.97545=0.34329\ amu$$

Energy released for $$1$$ atom of $$Ca = 319.6\ MeV$$
Energy release for $$1\ gm$$ of $$Ca = \cfrac{1}{39.975} \times 6.023 \times 10^{23} \times 319.6$$$$ = 4.81 \times 10^{24} MeV$$
Question 8
1 / -0

Calculate the binding energy of a deuteron atom, which consists of a proton and a neutron, given that the atomic mass of the deuteron is 2.014102 u

A
0.002388 MeV

B
2.014102 MeV

C
2.16490 MeV

D
2.224 MeV

Solution

$$ \begin{array}{l} \text { atomic mass M(H) of hydorgen and nuclear } \\ \text { mass }\left(m_{n}\right) \text { are } \\ m(H)=1.007825 u \text { and } M_n=1.008665u \\ \text { mass defect } \\ \Delta m=[M(H)+M_n-M(D)] \end{array} $$
$$ \begin{aligned} & 2.016490u-2.014102u \\ \Delta m &=0.002388u \end{aligned} $$
$$ \begin{array}{l} \text { As 1u correspond to 931.49 Mev energy } \\ \text { therefore mass defect corresponds to } \end{array} $$
$$ \begin{aligned} \text { energy } E_{b} &=0.002388 \times 931.49 \\ &=2.224 \mathrm{MeV} \\ \text { hence option } D & \text { is correct } \end{aligned} $$
Question 9
1 / -0

A stationary thorium nucleus $$(A=220, Z=90)$$ emits an alpha particle with kinetic energy $$E_{\alpha}$$. What is the kinetic energy of the recoiling nucleus?

A
$$\dfrac {E_{\alpha}}{108}$$

B
$$\dfrac {E_{\alpha}}{110}$$

C
$$\dfrac {E_{\alpha}}{55}$$

D
$$\dfrac {E_{\alpha}}{54}$$

Solution

$$ ^{220} Z _{90} \rightarrow \alpha + X$$
$$m_X = 220 - m_{\alpha} = 216$$

$$ E =\cfrac{p^2}{2m}$$
$$ p = \sqrt{2mE}$$

Momentum Conservation:
$$ \sqrt{2m_{\alpha}E_{\alpha}}= \sqrt{2m_{X}E_{X}}$$
$$m_{\alpha}E_{\alpha}= m_{X}E_{X}$$
$$ 4E_{\alpha}= 216E_{X}$$
$$ E(X) = \cfrac{E_{\alpha}}{54}$$
Question 10
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Directions For Questions

The compound unstable nucleus $$_{92}^{236}U$$ often decays in accordance with the following reaction:
$$_{92}^{236}U\rightarrow _{54}^{140}Xe+_{38}^{94}Sr+$$ other particles
During the reaction, the Uranium nucleus "fissions" (splits) into the two smaller nucleii. The reaction is energetically favorable because the smaller nucleii have higher nuclear binding energy per nucleon (although the lighter nuclei have lower total nuclear binding energies, because they contain fewer nucleons).
Inside a nucleus, the nucleons (protons and neutrons) attract each other with a "strong nuclear" force. All nucleons exert approximately the same strong nuclear force on each other. This force holds the nucleus together. Importantly, the strong nuclear force becomes important only when the protons and neutrons are very close together at intranuclear distances.

...view full instructions

In the nuclear reaction presented above, the "other particles" might be

A
an alpha particle, which consists of two protons and neutrons

B
two protons

C
one protons and one neutron

D
two neutrons

Solution

Difference in Mass number is $$236-140-94=2$$ and atomic number is $$92-38-54=0$$
which is most probably two neutrons.

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Nuclei Test - 61

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Nuclei Test - 61

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Question 1

emits an X-ray photon of energy $$2$$ MeV

captures an X-ray photon of energy $$2$$ MeV

emits an X-ray photon of energy $$ 3$$ MeV

captures an X-ray photon of energy $$3$$ MeV

Solution

Question 2

$$2.01 u$$

$$3.031 u$$

$$1.017 u$$

$$0.031 u$$

Solution

Question 3

0.001 g

1 g

10.0 g

1000 g

Solution

Question 4

$$75 MeV$$

$$6.5 MeV$$

$$1 eV$$

zero

Solution

Question 5

nitrogen of mass $$16$$

nitrogen of mass $$17$$

oxygen of mass $$16$$

oxygen of mass $$17$$

Solution

Question 6

$$13.9 MeV$$

$$26.9 MeV$$

$$23.6 MeV$$

$$19.2 MeV$$

Solution

Question 7

$$4.813\times 10^{24} MeV$$

$$4.813\times 10^{24} eV$$

$$4.813\times 10^{23} MeV$$

none of the above

Solution

Question 8

0.002388 MeV

2.014102 MeV

2.16490 MeV

2.224 MeV

Solution

Question 9

$$\dfrac {E_{\alpha}}{108}$$

$$\dfrac {E_{\alpha}}{110}$$

$$\dfrac {E_{\alpha}}{55}$$

$$\dfrac {E_{\alpha}}{54}$$

Solution

Question 10

an alpha particle, which consists of two protons and neutrons

two protons

one protons and one neutron

two neutrons

Solution

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