MCQ Questions for Class 12 Medical Physics Nuclei Quiz 3

Identify which of the following statement best describe a nuclear reaction.

0%
the particles in the nucleus are changed, and one element is transformed into another element when particles in the nucleus are gained or lost.
0%
the electrons are exchanged from one or more substances to produce a different substance, and the elements are the same in the products and reactants.
0%
two smaller nuclei are combined into a more massive nuclei giving off tremendous amounts of energy in the process.
0%
the nucleus of a large atom is split into two or more fragments.
0%
radiation is produced.

The energy needed to break a nucleus into its individual nucleons in a nuclear reaction is called

0%
nuclear binding energy
0%
ionization energy
0%
activation energy
0%
free energy
0%
fission energy

Which of the following describe that two lighter atoms combine to form one heavier atom?

0%
nuclear fission
0%
radioactive tracer
0%
nuclear fusion
0%
radiation therapy
0%
radioactive carbon dating

Which of the following term means the difference between the mass of the nucleus and the sum of each individual nucleon ?

0%
mass defect
0%
nucleon difference
0%
nucleon defect
0%
mass difference
0%
nuclear difference

Explanation

Formation of nucleus :

$Zp + (A-Z) n \rightarrow ^A_ZX$

Mass defect,

$\Delta M = [Zm_p + (A-Z)m_n] - M_{A,Z}$

Thus mass defect is defined as the difference between the mass of nucleus and the sum of each individual nucleon.

Substances which have identical chemical properties but differ in atomic weight are called.

0%
Isothermals
0%
Isotopes
0%
Isotropics
0%
Elementary particles

The rest energy involved in a mass of one atomic mass unit is _________ eV.

0%
$931$ MeV
0%
$1.6$ eV
0%
$9.3$ MeV
0%
$9.1$

Nuclear fusion occur in.

0%
Atom bomb
0%
Hydrogen bomb
0%
Neutron bomb
0%
None of these

Which of the following reaction must be initiated by the neutron?

0%
nuclear fission
0%
nuclear fusion
0%
radioactive carbon dating
0%
radioactive tracer
0%
radiation therapy

In the nuclear reaction pictured below, what is the missing nuclide?

$_{1}^{2}H + 7^{14}N \rightarrow _{2}^{3}He +?$

0%
$_{7}^{13}N$
0%
$_{6}^{13}C$
0%
$_{7}^{14}N$
0%
$_{6}^{11}C$
0%
$_{6}^{12}C$

Explanation

The given reaction :

$^{2}_{1} H$

$+$

$^{14}_{7} N$

$\rightarrow$

$^{3}_{2} He$

$+$

$^{A}_{Z} X$

Conservation of mass number :

$2 +14 = 3 + A$

$\implies A = 13$

Conservation of atomic number :

$1+7 = 2 + Z$

$\implies Z = 6$

Hence, the missing nuclide is

$^{13}_{6}C$ .

Q value for neutron decay is:

0%
$0.782\ MeV$
0%
$0.782\ eV$
0%
$78.2\ MeV$
0%
$0$

Explanation

For neutron decay, some mass disappears as neutrons convert to a proton, electron and antineutrino.

$Q=(m_n-m_p-m_{\bar{\nu}}-m_e)c^2$

$=0.782MeV$

A nuclear reaction with positive Q value is:

0%
endothermic
0%
exothermic
0%
either endothermic or exothermic
0%
neither endothermic nor exothermic

The average binding energy of the nucleon is

0%
$931$ MeV
0%
$8.5$ MeV
0%
$1.6$
0%
$3$

In beta decay, the typical Q value is approximately:

0%
$2\ MeV$
0%
$1\ MeV$
0%
$1\ eV$
0%
$10\ MeV$

Explanation

The

$Q$ value for a reaction is the amount of energy released by that reaction.

In beta decay, a typical

$Q$ is around

$1\ MeV$ .

The difference between the mass of a nucleus and the total mass of the constituents is its.

0%
packing fraction
0%
Mass defect
0%
Atomic mass
0%
None

When the nucleus of a radioactive element emits an alpha particle, the atomic number is decreased by

0%
$4$
0%
$2$
0%
$1$
0%
Zero

Binding energy of a nucleus is of the order of.

0%
Electron volt (eV)
0%
Kilo electron volt (KeV)
0%
Mega electron volt (MeV)
0%
A joule (J)

The correct equation of nuclear fusion reaction is

0%
$_{1}H^{1} + \,_{1}H^{1} \rightarrow H_{2}$
0%
$_{1}H^{2} +\, _{1}H^{1} \rightarrow\, _{2}He^{4} +\, _{0}n^{1}$
0%
$_{1}H^{2} +\, _{1}H^{2} \rightarrow\, _{2}He^{4} + energy$
0%
$2H_{2} +\, 3H_{2} \rightarrow 5H_{2}$

Explanation

In nuclear fusion reactions, the heavy isotopes of Hydrogen take part like deuterium

$_1H^2$ and tritium

$_1H^3$ as they have extra neutron. Therefore, the reaction C is correct out of the given options.

The nuclei having same number of protons but different number of neutrons are called _________.

0%
Isobars
0%
$\alpha$ -particles
0%
Isotopes
0%
$\gamma$ -particles

When a radioactive substance is kept in a vessel, the atmosphere around it is rich with

0%
$Ne$
0%
$Ar$
0%
$Xe$
0%
$He$

Explanation

A radioactive substance will emit

$\alpha$ radiation and alpha radiation is nothing but

$nucleus$ of

$Helium$ atom so the atmosphere will be rich with

$He$

Option D is correct.

All other options are the example of inert gases and any radioactive material doesn't emit inert gases.

$M_O$ is the mass of an oxygen isotope

$_8O^{17}$ ,

$M_p$ and

$M_n$ are the masses of a proton and a neutron, respectively, the nuclear binding energy of the isotope is

0%
$M_O c^2$
0%
$(M_O - 17 M_n)c^2$
0%
$(M_O - 8M_p)c^2$
0%
$(8M_p + 9M_n - M_O)c^2$

Explanation

Nuclear binding energy of a nucleus is given by ,

$B.E=\Delta mc^{2}=$ (mass of nucleons-mass of nucleus)

$c^{2}$

where

$\Delta m=$ mass defect

Number of protons in

$_{8}O^{17}$ ,

$p=8$

Number of neutrons in

$_{8}O^{17}$ ,

$n=17-8=9$

Hence , binding energy of

$_{8}O^{17}$ ,

$B.E.=(8M_{p}+9M_{n}-M_{O})c^{2}$

Which row describes the nature of

$\alpha$ - particles and of

$\gamma$ - rays

0%
$\alpha$ - particles : helium nuclei ; $\gamma$ rays - electromagnetic radiation
0%
$\alpha$ - particles : helium nuclei ; $\gamma$ rays - electrons
0%
$\alpha$ - particles : protons; $\gamma$ rays - electromagnetic radiation
0%
$\alpha$ - particles : protons; $\gamma$ rays - electrons

Explanation

Characteristics of

$\alpha$ particle

mass=

$4$ unit

charge =

$+2$ unit

The mass of Helium nuclei is

$4$ unit and charge is

$+2$ unit. So, it is equivalent to

$\alpha$ particle .

$\gamma$ - rays is a penetrating electromagnetic radiation arising from the radioactive decay of atomic nuclei.

$N$ atoms of a radioactive element emit

$n$ number

$\alpha$ - particle per second. Mean life of the elements

in second is :

0%
$\dfrac{n}{N}$
0%
$\dfrac{N}{n}$
0%
0693 $\dfrac{N}{n}$
0%
0.693 $\dfrac{n}{N}$

A nucleus of element X is represented as

$_{ 26 }^{ 56 }{ X }$
Which is an isotope of element X?

0%
$_{ 56 }^{ 26 }{ X }$
0%
$_{ 26 }^{ 54 }{ X }$
0%
$_{ 24 }^{ 56 }{ X }$
0%
$_{ 28 }^{ 54 }{ X }$

Explanation

Isotopes are variants of a particular chemical element which differ in neutron number, and consequently in nucleon number. All isotopes of a given element have the same number of protons but different numbers of neutrons in each atom.

Since , the isotopes has same number of proton , so , Option (B) is correct as it has same number of proton i.e.

$26$

The binding energy per nucleon of iron atom is approximately.

0%
13.6 eV
0%
8.8 MeV
0%
infinity
0%
10 MeV

Explanation

The maximum binding energy per nucleon occurs at around mass number $A=50$ , and corresponds to the most stable nuclei. Iron nucleusis located close to the peak with a binding energy per nucleon value of approximately $8.8\,MeV$ . It’s one of the most stable nuclides that exist.

1062109_1175912_ans_c2a81d7ccfb2401db9f74211c81c2c32.jpg

Which source of energy is used in a nuclear power station to generate electrical energy?

0%
Different types of atom regrouping
0%
Heavy nuclei splitting
0%
Radioactive isotopes decaying
0%
Radioactive atoms emitting $\beta$ -particles

The neutron was discovered by

0%
Marie Curie
0%
Pierree Curie
0%
James Chadwick
0%
Rutherford

The energy required to remove one neutron from

$_{13}Al^{27}$ is
(given mass of

$_{13}Al^{27}=$ 26.981541 amu, mass of

$_{13}Al^{26}=$ 25.984895 amu, mass of neutron

$=$ 1.008665 amu)

0%
6.525MeV
0%
11.195MeV
0%
4.232MeV
0%
8.464MeV

Explanation

$^{27}_{13}Al \rightarrow ^{23}_{13}Al + ^1n$

$BE = [25.984895 + 1.008665 - 26.981541][931.478]\ MeV$

$= 11.195\ MeV$

A proton and a neutron combine to give a deuterium nucleus.If

$m_{o}$ and

$m_{p}$ be the mass of neutron and proton respectively, then mass of deuterium nucleus is

0%
equal to $m_{o}$ + $m_{p}$
0%
more than $m_{o}$ + $m_{p}$
0%
less than $m_{o}$ + $m_{p}$
0%
can be less than or more than $m_{o}$ + $m_{p}$

Explanation

The energy released during this during this in form of gamma photon comes from mass defect. (i.e.,

$E=mc^2$ , where

$m$ is the mass defect). The mass of the deuterium nucleus (2.01355 u) is less than the sum of the masses of the proton (1.00728 u) and the neutron (1.00866 u), which is 2.01594 u.

In a nuclear reaction some mass converts into energy. In this reaction total B.E of reactants when compared with that of products is:

0%
always greater
0%
always less
0%
either greater or less
0%
always equal

$M, M$

$_{n}$ and

$M$

$_{p}$ denotes the masses of a nucleus of

$_{Z}X^{A}$ , a neutron, and a proton respectively. If the nucleus is separated into its individual protons and neutrons then,

0%
$M=(A-Z)M_{n}+ZM_{p}$
0%
$M=ZM_{n}+(A-Z)M_{p}$
0%
$M>(A-Z)M_{n}+ZM_{p}$
0%
$M<(A-Z)M_{n}+ZM_{p}$

Explanation

$Mc^{2} + Binding energy = [( A-Z)Mn + Z Mp] c^{2}$
Therefore, mass of nucleus is less than total mass of its free nucleons.

The nucleus finally formed in fusion of the proton in a proton cycle is that of:

0%
Helium
0%
Deuterium
0%
Carbon
0%
Hydrogen

Explanation

Proton - Proton cycle :
The thermonuclear reactions involved are :

$2(_1^1H) + 2(_1^1H) \Rightarrow 2(_1^2H) + 2(_1^0e) + Q_1$

$2(_1^2H) + 2(_1^1H) \Rightarrow 2(_2^3He) + Q_2$

$2(_2^3He) \Rightarrow (_2^4He) + 2(_1^2H) + Q_3$
On adding up these reactions, we obtain

$4(_1^1H) \Rightarrow _2^4He + 2(_1^0e) + Q$
where,

$Q = Q_1 + Q_2 + Q_3$ is the total energy involved in the fusion of 4 hydrogen nuclei to form Helium nucleus. The value of Q as calculated from mass defect comes out to be 26.7 MeV.

The binding energies of the atoms of elements

$P$ and

$Q$ are

$E_{p}$ and

$E_{Q}$ , respectively. Three atoms of element

$Q$ fuse to form one atom of element

$P$ . In this process, the energy released is

$e$ . The correct relation between

$E_{P}, E_{Q}$ and

$e$ will be

0%
$E_{Q}=3E_{P}+e$
0%
$E_{Q}=3E_{P} -e$
0%
$E_{P}=3E_{Q} +e$
0%
$E_{P}=3E_{Q}-e$

Explanation

Binding energy of

$3$

$Q$ elements will be

$3E_Q$
So, by energy conservation, binding energy of three

$P$ elements is equal to the sum of

$Q$ element's binding energy and the released energy.
So, equation is

$E_p + e = 3E_Q$

$E_p = 3E_Q - e$

In nuclear reaction

$_{4}Be^{9}+_{2}He^{4} \rightarrow _{6}c^{12}$ +X, X will be :

0%
Proton
0%
Neutron
0%
$\beta$ -particle
0%
$\alpha$ -particle

Explanation

$_{4}Be^{9} + _{2}He^{4} \rightarrow _{6}C^{12} + X$

No. of protons are already balanced with

$C$ .

$\therefore$

$X$ carries no charge.
and

$9+4 > 12$ by

$1$ unit mass.

$\therefore$

$X$ carries one unit mass.
It's a neutron.

Among the following reactions , the impossible one is :

0%
$^2He_4+$ $^4Be_9\longrightarrow$ $^0n_1+$ $^6C_{12}$
0%
$^2He_4 +$ $^7N_{14} \longrightarrow$ $^1H_1+$ $^8O_{17}$
0%
$4(^1H_1) \longrightarrow$ $^2He_4+$ $^2(^{-1}e_0)$
0%
$^3Li_7+$ $^1H_1\longrightarrow$ $^4Be_8$

The particle A is converted to C via following reactions then :

$A \rightarrow B + _{2}He^{4}$

$B \rightarrow C + 2 _{-1}e^{0}$

0%
A and C are isobars
0%
A and C are isotopes
0%
A and B are isobars
0%
A and B are isotopes

Explanation

No of neutrons

$= N_A N_B N_C$

No of protons

$= Z_A Z_B Z_C$

Mass

$= N_A+Z_A N_B+Z_B N_C+Z_C$

For reaction

$A \rightarrow B + _{2}He^{4}$

$Z_A = Z_B + 2 ............. [1]$

For reaction

$B \rightarrow C + 2 _{-1}e^{0}$

$Z_B = Z_C - 2 ..............[2]$

Solving these two equations, we get:

$Z_A = Z_C$

Hence they are isotopes.

On the bombardment of Boron with neutron, an

$\alpha$ - particle is emitted and product nucleus formed is

$\underline{\hspace{0.5in}}$

0%
$_{6}C^{12}$
0%
$_{2}Li^{6}$
0%
$_{3}Li^{7}$
0%
$_{4}Be^{9}$

Explanation

$_{5}B^{10}+^{1}n \rightarrow _{Z}^{A}X + _{2}^{4}He$

Mass balance:

$10 + 1 = A + 4$
Atomic no balance:

$5 + 0 = Z + 2$

$Z = 3$
Therefore, it's

$_{3}Li^{7}$

The missing particle in the reaction :

$^{253}_{99}Es+^{4}_{2}He\rightarrow ^{256}_{101}Md+ \underline{ \hspace{0.5in}}$

0%
deuteron
0%
proton
0%
neutron
0%
$\beta$ - particle

Explanation

Let the particle be

$X$ .
On summing the mass number on left hand side we get

$253+4 = 257$ units.
On right hand side we have

$256$ units only. So, we should have

$1$ unit mass number on

$X$ element.
On summing the atomic number on left hand side we get

$99 + 2 = 101$ units
On right hand side we have

$101$ units of atomic number. So, we should have

$0$ unit atomic number on

$X$ element.
So

$^1_0X = ^1_0n$ i.e. neutron.

Two deuterons combine to form a tritium nucleus and a __

0%
$\alpha$ -particle
0%
Fast neutron
0%
Proton
0%
$\beta$ -particle

Explanation

The given nuclear reaction is-

$^2_1H + ^2_1H \rightarrow ^3_1H + ^A_ZX$

where

$A$ and

$Z$ are the mass number and atomic number of element

$X$ , respectively.

Conservation of mass number :

$2+2 = 3+ A$

$\implies$

$A =1$

Conservation of atomic number :

$1+1 = 1+ Z$

$\implies$

$Z =1$

Thus the element

$X$ is a proton i.e.

$^1_1H$ .

The BE/A (binding energy per atomic mass) for deuteron and an

$\alpha$ particle are X

$_{1}$ and X

$_{2}$ respectively. The energy released in the reaction will be :

0%
X $_{2}$ -X $_{1}$
0%
2(X $_{2}$ -X $_{1}$ )
0%
4(X $_{2}$ -X $_{1}$ )
0%
8(X $_{2}$ -X $_{1}$ )

Explanation

$\dfrac{BE}{A}]_{_{1}^{2}D} = X_{1} , \dfrac{BE}{A}]_{\alpha} = X_{2}$

${BE}]_{D} = 2X_{1}$

${BE}]_\alpha = 4X_{2}$

$2 _{1}^{2}D \rightarrow _{2}^{4}He$

Energy released

$= BE]_{product} - BE]_{reactant}$

$= 4X_{2} - 2(2X_{1})$

$= 4(X_{2}-X_{1})$

The mass defect and binding energy per nucleon of an alpha particle are :
(Mp

$=$ 1.00734

$u$ , Mn

$=$ 1.00874

$u$ , M

$=$ 4.0015

$u$ )

0%
$0.0303u, 28.3 MeV$
0%
$0.1303 u, 28.3 MeV$
0%
$0.0303u, 7.07 MeV$
0%
$0.0306u, 7.14 MeV$

Explanation

$^4_2He$

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

$= 0.03066 u$

$B.E = 0.03066\times 931.478 MeV$

$= 28.56 MeV$

$B.E \ per\ nucleon = \dfrac{28.56}{4}$

$= {7.14 MeV}$

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

0%
positron
0%
two electrons
0%
two positrons
0%
an electron

Explanation

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.

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

0%
4,4
0%
4,2
0%
2,4
0%
4,6

Explanation

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.

Energy in the sun is due to

0%
Fossil fuels
0%
Radioactivity
0%
Fission
0%
Fusion

Mass of proton

$= 1.00760 amu$ , mass of neutron

$= 1.00899 amu$ , mass of deuterium nucleus

$=2.0147 amu$ . Then binding energy is :

0%
$0.00189 MeV$
0%
$1.76 amu$
0%
$1.76 MeV$
0%
$10^{15} joules$

Explanation

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

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

$= 0.00189 \mu$

$E = 0.00189\times 931.478 MeV$

$= 1.76 MeV$

The mass defect for the nucleus of Helium is

$0.0303 \ amu$ . The binding energy per nucleon in

$MeV$ is

0%
28
0%
7
0%
4
0%
1

Explanation

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

$= 28.22 MeV$

$B.E\ per\ nucleon = \dfrac{BE}{A} = \dfrac{28.22}{4} = 7.05 MeV$

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 :

0%
$283 J$
0%
$931.5 MeV$
0%
$28.3 MeV$
0%
$7.08 MeV$

Explanation

$2$ 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

$BE = \dfrac{28.32}{4} = 7.08\ MeV$

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)

0%
$3.49 MeV$
0%
$4.31MeV$
0%
$6.49 MeV$
0%
$931.49 MeV$

Explanation

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,

$1 amu = 931.49 MeV/c^2$

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

$931.5 MeV$ )

0%
$6.55 MeV$
0%
$7.55 MeV$
0%
$8.55 MeV$
0%
$9.55 MeV$

Explanation

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

$= 0.367211 \mu$

$BE = 0.367211\times 931.478 MeV$

$= 342.045 MeV$

BE per nucleon

$= \dfrac{BE}{40} = 8.55\ MeV$

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)

0%
$4.6 MeV$
0%
$5.8 MeV$
0%
$6.5 MeV$
0%
$8.2 MeV$

Explanation

$Cl$ there are

$17$ protons and

$(35 - 17)$

$=$ 18 neutrons.

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

$= 0.308995$

$BE = 0.308995\times 931.478 MeV$

$= 287.82 MeV$

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

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

0%
0.00657 amu
0%
0.00621 amu
0%
0.0657 amu
0%
0.6877 amu

Explanation

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.

$\Delta m = [2(1.00727) + 1(1.008665)] - 3.01664$

$\Delta m = [2.01454 + 1.008665] - 3.01664$

$\Delta m = 3.023205 - 3.01664$

$\Delta m = 0.006565 amu$

$\Delta m \approx 0.00657 amu$

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 3 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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