Nuclei Test

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Nuclei Test - 44

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Weekly Quiz Competition

Question 1
1 / -0

The binding energy per nucleon of $$_{ 8 }{ { O }^{ 16 } }$$ is $$7.97 MeV$$ and that of $$_{ 8 }{ { O }^{ 17 } }$$ is $$7.75 MeV$$. The energy required to remove one neutron from $$_{ 8 }{ { O }^{ 17 } }$$ is ___________ $$MeV$$.

A
$$3.62$$

B
$$3.52$$

C
$$4.23$$

D
$$7.86$$

Solution

$$_{ 8 }{ { O }^{ 16 } }\longrightarrow $$ $$_{ 8 }{ { O }^{ 17 } }+_{ 0 }{ { n }^{ 1 } }$$
∴ energy to remove neutron
= binding energy of $$_{ 8 }{ { O }^{ 17 }}$$ –binding energy of $$_{ 8 }{ { O }^{ 16 }}$$ = (17 × 7.75) – (16 × 7.79) = 4.23 MeV.
Question 2
1 / -0

What is the $$Q-value$$ of the reaction?
$$P + \,^{7}Li \rightarrow \, ^{4}He +\, ^{4}He$$
The atomic masses of $$^{1}H, ^{4}He$$ and $$^{7}Li$$ are $$1.007825u, 4.0026034u$$ and $$7.016004u$$ respectively.

A
$$17.35\ MeV$$

B
$$18.06\ MeV$$

C
$$177.35\ MeV$$

D
$$170.35\ MeV$$

Solution

The total mass of the initial particles
$$m_{i} = 1.007825 + 7.016004$$
$$= 8.023829u$$
and the total mass of final particles
$$m_{f} = 2\times 4.002603$$
$$= 8.005206u$$
Difference between initial and final mass of particles
$$\triangle m = m_{i} - m_{f}$$
$$= 8.023829 - 8.005206$$
$$= 0.018623u$$
The $$Q-value$$ is given by
$$Q = (\triangle m)c^{2}$$
$$= 0.018623\times 931.5$$
$$= 17.35\ MeV$$.
Question 3
1 / -0

Consider the nuclear reaction
$$X^{200}\rightarrow\,A^{110}+B^{90}+ energy $$
If the binding energy per nucleon for $$X , \,A$$ and $$ B$$ is $$7.4 MeV , 8.2 MeV$$ and $$8.2 MeV$$ respectively . What is the energy released?

A
$$200 MeV$$

B
$$160 MeV$$

C
$$110 MeV$$

D
$$90 MeV$$

Solution

Energy released $$=(E_A+E_B)-E_X$$
$$=(110\times 8.2+90 \times 8.2)-200 \times 7.4 $$
$$=1640-1480$$
$$=160 MeV$$
Question 4
1 / -0

$$^{234}U$$ has $$92$$ protons and $$234$$ nucleons total in its nucleus. It decays by emitting an alpha particle. After the decay it becomes.

A
$$^{232}U$$

B
$$^{232}Pa$$

C
$$^{230}Th$$

D
$$^{230}Ra$$

Solution

$$\displaystyle ^{234}_{92}U\overset{a}{\rightarrow} {^{230}_{90}Th}+\alpha$$
The mass number of thorium is $$230$$ and its atomic number, Z is $$90$$.
The mass number of radium is $$226$$ and it atomic number is $$88$$. As it is given $$^{230}Ra$$, students can make a mistake. Also the value of Z is not given. As at one instant, only one $$\alpha$$ particle can be emitted, unless it is given two successive emission of $$\alpha$$ particle, Ra cannot be obtained.
Question 5
1 / -0

A nuclear reaction along with the masses of the particles taking part in it is as follows:
$$A+B\longrightarrow C+D+QMeV$$
$$\underset { amu }{ 1.002 } +\underset { amu }{ 1.004 } \longrightarrow \underset { amu }{ 1.001 } +\underset { amu }{ 1.003 } +Q$$
The energy $$Q$$ liberated in the reaction is

A
$$1.234MeV$$

B
$$1.862MeV$$

C
$$0.931MeV$$

D
$$0.800MeV$$

Solution

by relation
$$Q=\sum { { m }_{ r } } -\sum { { m }_{ p } } =2.006-2.004$$
$$\therefore Q=0.002\times 931=1.86MeV\quad $$
Question 6
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Consider the following statements ($$X$$ and $$Y$$ stand for two different elements)
(I) $$_{32}X^{65}$$ and $$_{33}Y^{65}$$ are isotopes.
(II) $$_{42}X^{86}$$ and $$_{42}Y^{85}$$ are isotopes.
(III) $$_{85}X^{174}$$ and $$_{88}Y^{177}$$ have the same number of neutrons.
(IV) $$_{92}X^{235}$$ and $$_{94}Y^{235}$$ are isobars.
The correct statements are

A
II and IV only

B
I, II and IV only

C
II, III and IV only

D
I, II, III and IV

Solution

Two atoms are said to be isotope if they have same atomic number but different mass number.
So, $$_{42}X^{86}$$ and $$_{42}Y^{85}$$ are isotopes.
Number of neutrons in $$_{85}X^{174}$$ $$ = 174 - 85 = 89$$
Number of neutrons in $$_{88}Y^{177}$$ $$ = 174 - 85 = 89$$
So both have same number of neutrons.
Two atoms are said to be isobars if they have different atomic number but same mass number.
So, $$_{92}X^{235}$$ and $$_{94}Y^{235}$$ are isobars.
Question 7
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If M(A, Z), $$M_p$$ and $$M_n$$ denote the masses of the nucleus, proton and neutron respectively in units of $$u(1u=931.5 MeV/c^2)$$ and BE represents its binding energy in MeV, then.

A
$$M(A, Z)=ZM_p+(A-Z)M_n-BE$$

B
$$M(A, Z)=ZM_p+(A-Z)M_n+BE/c^2$$

C
$$M(A, Z)=ZM_p+(A-Z)M_n-BE/c^2$$

D
$$M(A, Z)=AM_p+(A-Z)M_n+BE$$

Solution

$$ZM_p +(A-Z)M_n= M(A,Z) $$
Mass effect= $$\dfrac{B.E.}{c^2}$$
$$M(A, Z)=ZM_p+(A-Z)M_n-BE/c^2$$
Question 8
1 / -0

Find the binding energy of an electron in the ground state of a hydrogen like atom in whose spectrum the third of the corresponding Balmer series is equal to 108.5 nm

A
54.4 eV

B
13.6 eV

C
112.4 eV

D
None of these

Solution

Binding energy in the ground state

$$E_b=\frac{E_0 Z^2}{n^2}$$

$$=\frac{E_0Z^2}{1^2}$$

$$=E_0Z^2$$

Wave number of third Balmer line is

$$v=\frac{E_0 Z^2}{hc} (\frac{1}{2^2}-\frac{1}{5^2})$$

$$=\frac{E_0Z^2}{hc}. \frac{21}{100}$$

Wavelength of third Balmer line

$$\lambda=\frac{1}{v}$$

$$=\frac{hc}{E_0Z^2}. \frac{100}{21}$$

$$=\frac{19.86*10^{-26}}{E_b} . \frac{100}{21}$$

$$E_b=\frac{19.86*10^{-26}} {\lambda} . \frac{100}{21}$$

$$=\frac{19.86*10^{-26}}{108.5*10^{-9}} . \frac{100}{21}$$

$$=8.7$$ X $$10^-18 J$$

$$=\frac{8.7*10^{-18}}{1.6*10^-19} eV$$

$$=54.4 eV$$
Question 9
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Which of the following statements is correct?

A
The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons

B
The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons

C
In nuclear fission, energy is released by fusion two nuclei of medium mass(approximately $$100$$amu)

D
None of the above

Solution

The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons.
In nuclear fission, energy is released by fragmentation of a very low nucleus
Question 10
1 / -0

An energy of $$24.6eV$$ is required to remove one of the electrons from a neutral helium atom. The energy (in eV) required to remove both the electrons from a neutral helium atom is

A
$$38.2$$

B
$$49.2$$

C
$$51.8$$

D
$$79.0$$

Solution

When one electron from neutral helium atom is removed, it becomes hydrogen like. The ionisation energy of hydrogen like atom is given by
$${ E }_{ ion }={ Z }^{ 2 }Rhc$$
for helium, $$Z=2$$ $$Rhc=13.6eV$$
$$\therefore$$ Energy required to liberate second electron from helium
$$\quad { E }_{ 2 }={ E }_{ ion }={ 2 }^{ 2 }\times 13.6eV=-54eV$$
$$\therefore$$ Energy required to remove both the electrons from helium is
$$={ E }_{ 1 }+{ E }_{ 2 }=(24.6+54.4)eV=79.0eV$$

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Re-Attempt Weekly Quiz Competition

Nuclei Test - 44

Result

Nuclei Test - 44

Score

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SHARING IS CARING

Question 1

$$3.62$$

$$3.52$$

$$4.23$$

$$7.86$$

Solution

Question 2

$$17.35\ MeV$$

$$18.06\ MeV$$

$$177.35\ MeV$$

$$170.35\ MeV$$

Solution

Question 3

$$200 MeV$$

$$160 MeV$$

$$110 MeV$$

$$90 MeV$$

Solution

Question 4

$$^{232}U$$

$$^{232}Pa$$

$$^{230}Th$$

$$^{230}Ra$$

Solution

Question 5

$$1.234MeV$$

$$1.862MeV$$

$$0.931MeV$$

$$0.800MeV$$

Solution

Question 6

II and IV only

I, II and IV only

II, III and IV only

I, II, III and IV

Solution

Question 7

$$M(A, Z)=ZM_p+(A-Z)M_n-BE$$

$$M(A, Z)=ZM_p+(A-Z)M_n+BE/c^2$$

$$M(A, Z)=ZM_p+(A-Z)M_n-BE/c^2$$

$$M(A, Z)=AM_p+(A-Z)M_n+BE$$

Solution

Question 8

54.4 eV

13.6 eV

112.4 eV

None of these

Solution

Question 9

The rest mass of a stable nucleus is less than the sum of the rest masses of its separated nucleons

The rest mass of a stable nucleus is greater than the sum of the rest masses of its separated nucleons

In nuclear fission, energy is released by fusion two nuclei of medium mass(approximately $$100$$amu)

None of the above

Solution

Question 10

$$38.2$$

$$49.2$$

$$51.8$$

$$79.0$$

Solution

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