MCQ Questions for Class 12 Medical Physics Nuclei Quiz 6

From the following equations, pick out the possible nuclear reactions.

0%
${ { _{ 6 }{ C } } }^{ 13 }\quad +\quad { { _{ 1 }{ H } } }^{ 1 }\quad \rightarrow \quad { { _{ 6 }{ C } } }^{ 14 }\quad +\quad 4.3\quad MeV$
0%
${ { _{ 6 }{ C } } }^{ 12 }\quad +\quad { { _{ 1 }{ H } } }^{ 1 }\quad \rightarrow \quad { { _{ 7 }{ N } } }^{ 13 }\quad +\quad 2\quad MeV$
0%
${ { _{ 7 }{ N } } }^{ 14 }\quad +\quad { { _{ 1 }{ H } } }^{ 1 }\quad \rightarrow \quad { { _{ 8 }{ O } } }^{ 15 }\quad +\quad 7.3\quad MeV$
0%
${ { _{ 92 }U } }^{ 235 }\quad +\quad { { _{ 0 }{ n } } }^{ 1 }\quad \rightarrow \quad { { _{ 54 }{ X } } }^{ 140 }\quad +\quad { { _{ 38 }Si } }^{ 94 }\quad +\quad { { _{ 20 }n } }^{ 1 }\quad +\quad \gamma \quad +\quad 200\quad MeV$

Statement-I : Energy is released when heavy nuclei undergo fission or light nuclei undergo fusion and
Statement-11: For heavy nuclei, binding energy per nucleon increases with increasing Z. while for light nuclei it decreases with increasing Z.

0%
Both the statements are true and Statement 2 is the correct explanation for Statement 1.
0%
Both the statements are true but Statement 2 is not the correct explanation for Statement 1.
0%
Statement 1 is true but Statement 2 is false.
0%
Statement 1 is false but Statement 2 is true.

Which statement about nuclear fusion is false?

0%
Nuclear fusion happens in the sun
0%
Nuclear fusion is the joining of the nuclei of atoms
0%
Nuclear fusion is currently used to generate electrical energy
0%
Nuclear fusion can use hydrogen as fuel.

Explanation

Nuclear fusion reactions are not currently used to generate electricity because this process is not based on chain reactions as in nuclear fission. In fact, this process (fusion) requires very high temperature.

Thus option

$(C)$ is the answer.

Consider the nuclear change

$^{237}_{93}X\, \xrightarrow{- \alpha}\, A\, \xrightarrow{- \beta}\, B\xrightarrow{- \alpha}\,C$
Which of the following statements is not correct?

0%
Mass number of B is 233
0%
Atomic number of A is 91
0%
C is an isotope of A
0%
B is an isobar of A

Explanation

$_{93}X^{237} \xrightarrow{-\alpha} _{91}A^{233}\xrightarrow{-\beta} _{92}B^{233} \xrightarrow{-\alpha} _{90}C^{229}$

(Isotopes have same atomic number but different mass number)

Radioactivity is generally does not found in :

0%
light nuclei
0%
stable nuclei
0%
heavy nuclie
0%
nuclei of intermediate mass

Hydrogen has________ isotopes.

0%
2
0%
3
0%
4
0%
5

Explanation

Hydrogen has 3 isotopes namely. protium

$_1^1H$ , deuterium

$_1^2H$ and tritium

$_1^3H$

Modern atomic mass scale is based on the mass of:

0%
$H - 1$
0%
$C - 12$
0%
$C - 14$
0%
$C - 16$

Explanation

Modern atomic weight scale is based on

$C - 12$ .
The standard unit for expressing the mass of atom is amu (atomic mass unit).
It is equal to 1/12 of the mass of an atom of carbon-12 equal to

$1.6605 \times 10^{-19} \: g$ .
amu is also called as avogram.
Avogram is a unit of mass and weight equal to one gram divided by the Avogadro's number.

Which of the following statements about radioactivity is/are true ?

0%
It involves outer electrons activity
0%
It is not affected by temperature or pressure
0%
It is an exothermic process
0%
The radioactivity of an element is not affected by any other element compounded by it

Atomic mass of an element is:

0%
actual mass of one atom of the element
0%
average mass of an atom of different atoms of the element
0%
always a whole number
0%
None of these

Which of the following change the activity of radioisotope ?

0%
Temperature
0%
Pressure
0%
Chemical environment
0%
None of the above

Which of the following isotopes is likely to be most stable ?

0%
$^{71}_{30}Zn$
0%
$^{66}_{30}Zn$
0%
$^{40}_{20}Ca$
0%
None of these

As compared to

$^{12}C$ atom,

$^{14}C$ atom has

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Two extra protons and two extra electrons.
0%
Two extra protons and no extra electron.
0%
Two extra neutrons and no extra electron.
0%
Two extra neutrons and two extra electrons.

Explanation

Carbon

$-14$ or radiocarbon is a radioactive isotope of carbon with a nucleus containing

$6$ protons and

$8$ neutrons while

$C-12$ contains

$6$ protons and

$6$ neutrons.

The reaction

$^2_1H\, +\, ^2_1H\, \rightarrow\, ^3_2He\, +\, ^1_0n$ is called

0%
Fusion
0%
Fission
0%
Endothermic reaction
0%
Spontaneous reaction

Which of the following is an example of nuclear fusion reaction?

0%
$^{233}_{90}TH\, +\, ^1_0n\, \rightarrow\, ^{234}_{90}TH$
0%
$^{113}_{48}Cd\, +\, ^1_0n\, \rightarrow\, ^{114}_{48}Cd\, +\, \gamma\, -ray$
0%
$^3_1H\, +\, ^1_1H\, \rightarrow\, ^4_2He$
0%
$^{239}_{92}U\, \rightarrow\, ^{239}_{93}Np\, +\, ^0_{-1}e$

Which of the following is used in cancer treatment ?

0%
$^{131}_{53}I$
0%
$^{32}_{15}P$
0%
$^{60}_{27}Co$
0%
$^{2}_{1}H$

$^{23}Na$ is the more stable isotope of

$Na$ . By which process

$^{24}_{11}Na$ can undergo radioactive decay?

0%
$\beta^--emission$
0%
$\alpha-emission$
0%
$\beta^+-emission$
0%
None of these

Explanation

To convert

$^{24}_{11}Na$ to

$^{23}_{11} Na$ , we need to decrease the mass number of

$^{24}_{11}Na$ by

$1$ .

Since, in

$\beta^{-}$ emission and

$\beta^{+}$ emission, the mass number of product nucleus is same as that of parent nucleus.

Whereas, in

$\alpha$ emission, the mass number of product nucleus gets decreased by

$4$ compared to that of parent nucleus.

Thus

$^{24}_{11} Na$ cannot undergo any of the radioactive decay to convert into

$^{23}_{11}Na$ .

In the nuclear reaction

$^{14}_7N\, +\, _2^4He\, \rightarrow\, ^{17}_8O\, +\, ...............$
What is the missing particle ?

0%
$_2^4He$
0%
$_1^1H$
0%
$_0^1n$
0%
$\gamma$

Which bomb works on the principle of Nuclear fusion ?

0%
Atom bomb
0%
Hydrogen bomb
0%
Neutron bomb
0%
Nuclear reactor

Fusion reaction is also known as ..........

0%
Chemical reaction
0%
Elastic reaction
0%
Thermonuclear reaction
0%
Photonuclear reaction

Pick out the isoelectronic species from the following :

$(I)\, {CH_{3}}^{+}\, \quad\, (II)\, H_3O^+\, \quad\, (III)\, NH_3\, \quad\, (IV)\, {CH_{3}}^{-}$

0%
I and II
0%
III and IV only
0%
I and III
0%
II, III, IV

Biding energy is equal to: (m is the mass defect in amu)

0%
$\Delta m \times 931 MeV$
0%
$\Delta m \times 831 MeV$
0%
$\Delta m \times 1031 MeV$
0%
$\Delta m \times 731 MeV$

Explanation

From mass energy equivalence Binding energy,

$E=\triangle m{ c }^{ 2 }\\ \because { c }^{ 2 }=931MeV\\ E=\triangle m \times 931MeV$

When a hydrogen bomb explodes, which of the following is used?

0%
Fission
0%
Fussion
0%
Both
0%
None of these

A natural phenomenon that supports the experimental conclusion that atoms are divisible is :

0%
allotropy
0%
radioactivity
0%
cracking
0%
none of these

Two atoms of the same element are found to have different number of neutrons in their nuclei. These two atoms are :

0%
Isomers
0%
lsotopes
0%
Isobars
0%
Allotropes

Which atom contains exactly 15 neutrons?

0%
$P^{32}$ (atomic number $= 17$ )
0%
$S^{32}$ (atomic number $= 16$ )
0%
$O^{15}$ (atomic number $= 8$ )
0%
$N^{15}$ (atomic number $= 7$ )

Explanation

$mass\space number= no. of \space protons + no. of\space neutrons$
Given that mass number of P = 32

Atomic number (no. of electrons

$=$ no. of protons)

$=$ 17

Number of neutrons

$= 32 - 17 = 15$

Hence, P satisfies the requirement.

Mass number is denoted by:

0%
D
0%
S
0%
A
0%
Z

The difference between the sum of the masses of the constituent particles and the mass of an atom is called ___________.

0%
Mass number
0%
Mass defect
0%
Binding energy
0%
None of these

Explanation

The difference between the sum of the masses of the constituent particles and the mass of an atom is called mass defect.

The energy released due to mass defect was given by Einstein

$E=\triangle mc^2$

${ _{ 7 }{ { N }^{ 14 } }_{ } }$ is bombarded with an

$\left(\alpha\right)$ alfa particle to produce a proto and an unknown nucleus

$'X',$

${ _{ 7 }{ { N }^{ 14 } }_{ } } + \alpha$ (particle)

$\longrightarrow X + p$
Then

$'X'$ stands for

0%
${ _{ 7 }{ { N }^{ 15 } }_{ } }$
0%
${ _{ 8 }{ { N }^{ 14 } }_{ } }$
0%
${ _{ 8 }{ { O }^{ 17 } }_{ } }$
0%
${ _{ 8 }{ { O }^{ 16 } }_{ } }$

Explanation

The required reaction is

${ _{ 7 }{ { N }^{ 14 } }_{ } }+{ _{ 2 }{ { He }^{ 4 } }_{ } }\longrightarrow { _{ 8 }{ { O }^{ 17 } }_{ } }+{ _{ 1 }{ { H }^{ 1 } }_{ } }$

Artificial radioactivity was discovered by:

0%
Rontegen
0%
Becquerel
0%
Marie Curie
0%
Irene Joliot Curie

One atomic mass unit is equivalent to

0%
$1.67\times 10^{-27} g$
0%
$1.66\times 10^{-27} kg$
0%
$1.67\times 10^{-27} mg$
0%
$1.6\times 10^{-24} g$

Explanation

One amu

$=\frac {1}{N_A}=\frac {1}{6.02\times 10^{23}}$

$=1.66\times 10^{-24}g=1.66\times 10^{-27}kg$

So, the correct options are

$B$ and

$D$

What are the different isotopes formed when protons combine to form a

$_{2}{He}^{4}$ nucleus ?

0%
$_{1}{H}^{1}$ , $_{2}{He}^{3}$
0%
$_{2}{He}^{3}$ , $_{2}{He}^{3}$
0%
$_{1}{H}^{1}$ , $_{1}{He}^{1}$
0%
$_{2}{H}^{3}$ , $_{1}{H}^{2}$

Explanation

$_1^1H$ and

$_2^3He$ combine to form

$_2He^4$ nucleus.

$_{ 2 }^{ 3 }{ He }+_{ 1 }^{ 1 }{ H }\rightarrow _{ 2 }{ He }^{ 4 }+E$

An atom bomb works on :

0%
Electrical energy
0%
Heat energy
0%
Mechanical energy
0%
Atomic energy

What is the mass defect and the binding energy of

$_{27}{Co}^{59}$ which has a nucleus of mass of

$58.933 u$ ?

$\left({m}_{p} = 1.0078 u, {m}_{n} = 1.0087 u \right)$

0%
517.914 MeV
0%
617.914 MeV
0%
17.914 MeV
0%
717.914 MeV

Explanation

$_{27}{Co}^{59}$ number of protons

$= 27$ .
Number of neutrons

$= 59 - 27 = 32$ .

$\therefore$ The total mass of the nucleus

$= 27 \times 1.0078 u + 32 \times 1.0087 u = 27.2106 u + 32.2784 u = 59.489 u$ .
Given, the actual mass of the nucleus

$= 58.933 u$ .

$\therefore$ The mass defect

$= 59.489 u - 58.933 u = 0.556 u$ .

$\therefore$ The binding energy

$= \Delta m \times 931.5 MeV = 0.556 u \times 931.5 MeV = 517.914 MeV$ .

Find the binding energy per nucleon for lithium nucleus

$_{3}{Li}^{7}$ .

Take mass of

$_{3}{Li}^{7} = 7 u$ , mass of proton

$= 1.007825 u$ and mass of neutron

$= 1.008665 u$ .

Take

$1 u = 931.5 MeV$ .

0%
$3.7 MeV$
0%
$4.9 MeV$
0%
$1.7 MeV$
0%
$7.736 MeV$

Explanation

Binding energy per nucleon.

$E=\frac{\triangle E} {A}=\frac{\triangle m * 931}{A}$

$\triangle m= (3M_p+4M_n) -$ Mass of

$_3Li^7$

Where,

$M_p$ is the mass of the proton

$M_n$ is the mass of the neutron

$\triangle m=$

$(3$ X

$1.007825+4$ X

$1.008898)-7=0.059067$

$\triangle E=0.059067$ X

$931=54.991377$

$E=\frac{\triangle E}{A}=\frac{54.991377}{7}=7.736MeV$

Calculate the binding energy per nucleon for Beryllium

$_{4}{Be}^{9}$ , its mass being

$9.012 \ u$ . The masses of proton and neutron are

$1.008 u$ and

$1.009 u$ . (Take

$1 u = 931.5 MeV$ )

0%
8.3636 MeV
0%
7.2828 MeV
0%
9.4444 MeV
0%
6.7275 MeV

Explanation

First, let us find the mass defect. The Beryllium nucleus contains 4 protons and 5 neutrons.
Mass of 4 protons

$= 4 \times 1.008 = 4.032 u$ ; Mass of 5 neutrons

$= 5 \times 1.009 u = 5.045 u$
Total mass of protons (4) and neutrons (5)

$= 4.032 + 5.045 = 9.077 u$
Mass defect

$= 9.077 - 9.012 = 0.065 u$
The mass defect converted into equivalent energy gives binding energy.

$1 u = 931.5 MeV$

$\therefore 0.065 u = 0.065 \times 931.5 MeV = 60.5475 MeV$
Binding energy per nucleon

$= \displaystyle\frac{Binding energy}{Number of nucleons} = \displaystyle\frac{60.5475}{9} = 6.7275 MeV$

If the binding energy per nucleon for

$_{3}{Li}^{7}$ is

$5.6 MeV$ , the total binding energy of a lithium nucleus is?

0%
$139.2 MeV$
0%
$39.2 MeV$
0%
$15.8 MeV$
0%
$115.8 MeV$

Explanation

Binding energy per nucleon = 5.6 MeV

No. of nucleon = No. of proton + No. of neutron

=3 + 4 = 7

So, for 7 Nucleon

$=7\times 5.6 \\=39.2MeV$

Complete the following reaction.

$_{13}{Al}^{27} +$ __________

$\rightarrow _{15}{P}^{30} + _{0}{n}^{1}$

0%
$\alpha$
0%
$\beta$
0%
$\gamma$
0%
$_{ 1 }^{ 1 }{ H }$

Explanation

${ _{ 13 }{ Al }^{ 27 } }+{ _{ 2 }{ \alpha }^{ 4 } }\rightarrow { _{ 15 }{ P }^{ 30 } }+{ _{ 0 }{ n }^{ 1 } }$

Atomic No. : 13+2=15+0

Mass No. : 27+4=30+1

So,

$\alpha$ particle will be emitted.

Chlorine has ______isotopes.

0%
1 (one)
0%
2 (two)
0%
3 (three)
0%
4 (four)

Explanation

Chlorine has two isotopes

$_{17}^{35} Cl$ and

$_{17}^{37} Cl$ each isotope in its Nucleus has 17 protons.

What is the atomic mass (u) of chlorine?

0%
34
0%
35.5
0%
33
0%
35

Carbon has __________ isotopes.

0%
1 (one)
0%
2 (two)
0%
3 (three)
0%
4 (four)

Explanation

Carbon has three isotopes

$_6^{12} C, \ \ _6^{13} C \ \ and \ \ _6^{14} C$

$M_0$ 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%
$(M_O-8M_p)c^2$
0%
$(M_O-8M_p-9M_n)c^2$
0%
$M_Oc^2$
0%
$(M_O-17M_n)c^2$

Explanation

Number of proton in Oxygen isotope,

$Z =8$

Number of neutron,

$N = 17 - 8 = 9$

The nuclear reaction:

$8p$

$+$

$9 n$

$\rightarrow$

$_8O^{17}$

Nuclear binding energy of the isotope,

$B.E = (M_{product} - M_{Reactant}) c^2$

$\Rightarrow$

$B.E = (M_{O} - 8M_p - 9M_n) c^2$

Which scientist proposed the concept of atomic mass?

0%
Avogadro
0%
Gay Lussac
0%
Proust
0%
Dalton

${m}_{p}$ and

${m}_{n}$ are masses of proton and neutron respectively. An element of mass

$M$ has

$Z$ protons and

$N$ neutrons then:

0%
$M> Z{m}_{p}+N{m}_{n}$
0%
$M=Z{m}_{p}+N{m}_{n}$
0%
$M< Z{m}_{p}+N{m}_{n}$
0%
$M$ may be greater than less than or equal to $Z{m}_{p}+N{m}_{n}$ , depending on nature of element

Explanation

When a nucleus is formed, then the mass of nucleus is slightly less than the sum of the mass of

$Z$ protons and

$N$ neutrons
i.e.

$M< (Z{m}_{p}+N{m}_{n})$

${M}_{0}$ is the mass of an oxygen isotope

$_{8}{O}^{17}, {M}_{p}$ and

${M}_{n}$ are the masses of a proton and a neutron, respectively, then the nuclear binding energy of the isotope is

0%
$\left({M}_{o} - 8{M}_{p}\right) {c}^{2}$
0%
$\left({M}_{o} - 8{M}_{p} - 9{M}_{n}\right) {c}^{2}$
0%
${M}_{o}{c}^{2}$
0%
$\left({M}_{o} - 17{M}_{n}\right) {c}^{2}$

Explanation

$\begin{aligned}M_{0} &=\text{mass of $_8O^{17}$}\\M_{p} &=\text { Mass of proton } \mathrm{} \\M_{n} &=\text { Mass of neutron } \mathrm{}_{}\end{aligned}$

$\text { Binding energy }=\text { mass defect }(\Delta m)\times c^{2}$

$\begin{array}{l}\Delta m=\left(\text { Mass of Oxygen $-$ total mass of }{\text {protons }}\right.- \text {total mass of}\text { neutrons } \\\text { Binding energy }=\left(M_{0}-8M_p -9M_n\right) c^{2}\end{array}$

Fusion reaction takes place at high temperature because

0%
KE is high enough to overcome repulsion between nuclei
0%
Nuclei are most stable at this temperature
0%
Nuclei are unstable at this temperature
0%
None of the above

An alpha particle

$(_{ }^{ 4 }{ He })$ has a mass

$4.00300\ amu$ . A proton has a mass

$1.00783$ amu and a neutron has a mass

$1.00867\ amu$ respectively. The binding energy of alpha estimated from these data is closest to:

0%
$27.9\ MeV$
0%
$22.3\ MeV$
0%
$35.0\ MeV$
0%
$20.4\ MeV$

Explanation

The mass defect,

$\Delta m=2\left( { m }_{ p }+{ m }_{ n } \right) -{ m }_{ He }$

$=2(1.00783+1.00867)-4.00300$

$=0.300\ amu$
So, the binding energy

$E=\Delta m{ c }^{ 2 }$

$=0.03\times 931MeV=27.9\ MeV\quad$

One atomic mass is equal to

0%
$1.66\times10^{-27}$ g
0%
$1.66\times10^{-24}$ g
0%
$1.66\times10^{-23}$ g
0%
$1.66\times10^{-25}$ g

Explanation

One atomic mass unit is equal to

$1.66 \times 10^{-24} g.$ It is equal to 1/12th of mass of atom of Carbon-12. It is abbreviated as

$\text{u.}$

When three

$\alpha-particles$ combined to from a

$^{12}C$ nucleus, the mass defect is
(atomic mass of

$_{2}He^{4}$ is

$4.002603\ u$ )

0%
$0.007809\ u$
0%
$0.002603\ u$
0%
$4.002603\ u$
0%
$0.5\ u$

Explanation

Mass defect,

$\triangle =$ Total mass of

$\alpha-particles -$ mass of

$^{12}C$ nucleus

$= 3\times 4.002603 - 12$

$= 12.007809 - 12$

$= 0.007809\ unit$

${ F }_{ pp },{ F }_{ nn }$ and

${ F }_{ np }$ are the nuclear forces between proton-proton, neutron-neutron and neutron-proton respectively. Then relation between them is

0%
${ F }_{ pp }={ F }_{ nn }\neq { F }_{ np }$
0%
${ F }_{ pp }\neq { F }_{ nn }={ F }_{ np }$
0%
${ F }_{ pp }={ F }_{ nn }={ F }_{ np }$
0%
${ F }_{ pp }\neq { F }_{ nn }\neq { F }_{ np }$

Explanation

Nuclear force between two particles is independent of charges of particles.

$\Rightarrow { F }_{ pp }={ F }_{ nn }={ F }_{ np }$

Which of the following is a fusion reaction?

0%
$_{1}H^{2} + _{1}H^{2} \rightarrow _{2}He^{4}$
0%
$_{1}H^{2} + _{1}H^{2} \rightarrow 2_{1}He^{2}$
0%
$_{1}H^{1} + _{1}H^{1} \rightarrow _{2}He^{4}$
0%
$_{1}H^{1} + _{1}H^{2} \rightarrow _{2}He^{4} + n$

Explanation

$_{ 1 }{ H }^{ 2 }+_{ 1 }{ H }^{ 2 }\rightarrow _{ 2 }{ He }^{ 4 }$ is a fusion reaction because here two smaller nuclei fuse together to form a single stable nuclei.

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 6 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 6 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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