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
1 / -0

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
1 / -0

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
1 / -0

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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Nuclei Test - 44

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Nuclei Test - 44

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Question 1

3.623.623.62

3.523.523.52

4.234.234.23

7.867.867.86

Solution

Question 2

17.35 MeV17.35\ MeV17.35 MeV

18.06 MeV18.06\ MeV18.06 MeV

177.35 MeV177.35\ MeV177.35 MeV

170.35 MeV170.35\ MeV170.35 MeV

Solution

Question 3

200MeV200 MeV200MeV

160MeV160 MeV160MeV

110MeV110 MeV110MeV

90MeV90 MeV90MeV

Solution

Question 4

232U^{232}U232U

232Pa^{232}Pa232Pa

230Th^{230}Th230Th

230Ra^{230}Ra230Ra

Solution

Question 5

1.234MeV1.234MeV1.234MeV

1.862MeV1.862MeV1.862MeV

0.931MeV0.931MeV0.931MeV

0.800MeV0.800MeV0.800MeV

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)=ZMp+(A−Z)Mn−BEM(A, Z)=ZM_p+(A-Z)M_n-BEM(A,Z)=ZMp​+(A−Z)Mn​−BE

M(A,Z)=ZMp+(A−Z)Mn+BE/c2M(A, Z)=ZM_p+(A-Z)M_n+BE/c^2M(A,Z)=ZMp​+(A−Z)Mn​+BE/c2

M(A,Z)=ZMp+(A−Z)Mn−BE/c2M(A, Z)=ZM_p+(A-Z)M_n-BE/c^2M(A,Z)=ZMp​+(A−Z)Mn​−BE/c2

M(A,Z)=AMp+(A−Z)Mn+BEM(A, Z)=AM_p+(A-Z)M_n+BEM(A,Z)=AMp​+(A−Z)Mn​+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 100100100amu)

None of the above

Solution

Question 10

38.238.238.2

49.249.249.2

51.851.851.8

79.079.079.0

Solution

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$3.62$

$3.52$

$4.23$

$7.86$

$17.35\ MeV$

$18.06\ MeV$

$177.35\ MeV$

$170.35\ MeV$

$200 MeV$

$160 MeV$

$110 MeV$

$90 MeV$

$^{232}U$

$^{232}Pa$

$^{230}Th$

$^{230}Ra$

$1.234MeV$

$1.862MeV$

$0.931MeV$

$0.800MeV$

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

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

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

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

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

$38.2$

$49.2$

$51.8$

$79.0$