MCQ Questions for Class 12 Medical Physics Nuclei Quiz 13

Binding energy of deuterium is 2.23 MeV, then its mass defect in a.m.u. is:
The value of energy for

$_1H^2, _2He^4, _{26}Fe^{56}, _{92}U^{235}$ .

0%
-0.0024
0%
-0.0012
0%
0.0012
0%
0.0024

The binding energy of

$\alpha$ -particle

$\begin{array} { l } { 4 } \\ { 2 } \end{array}He$ is

$7.047$

$MeV$ per nucleon and the binding energy of deutron

$\begin{array} { l } { 2} \\ { 1 } \end{array}H$ is

$1.112$

$MeV$ per nucleon. Then in the fusion reaction

$_ { 1 } ^ { 2 } H + _ { 1 } ^ { 2 } H \rightarrow _ { 2 } ^ { 4 } H e + Q$ , the energy

$Q$ released is

0%
$23.74 \mathrm { MeV }$
0%
$32.82 \mathrm { MeV }$
0%
$11.9 MeV$
0%
$4.94 \mathrm { MeV }$

Two radioactive nuclei P and Q in a given sample decay into a stable nucleus R. At time t = 0, number of P species are

$4 N_o$ and that of Q are

$N_o$ . Half-life of P (for conversion to R) is 1 minute where as that of Q is 2 minutes. Initially there are no nuclei of R present in the sample. When number of nuclei of P and Q are equal, the number of nuclei of
R present in the sample would be :-

0%
$\dfrac{9 N_o}{2}$
0%
$\dfrac{ 7N_o}{2}$
0%
$\dfrac{N_o}{4}$
0%
$\dfrac{9 N_o}{4}$

A nucleus of atomic mass M emits a gamma ray photon of frequency v. then

0%
Kinetic energy acquired by nucleus $= \frac { h ^ { 2 } v ^ { 2 } } { 2 M c ^ { 2 } }$
0%
Loss in the internal energy of the nucleus $h v \left( 1 + \frac { h v } { 2 M c ^ { 2 } } \right)$
0%
Momentum of nucleus is equal and opposite to $\gamma$ ray photon
0%
Loss in internal energy of nucleus is hv

Explanation

Hence, option

$(A)$ is correct answer.
1380719_1360634_ans_b054f5bef76842e5b31fee65302e9470.jpg

The above is a plot of binding energy per nucleon

$E _ { b }$ , against the nuclear mass

$\mathrm { M } ; \mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } , \mathrm { E } , \mathrm { F }$ correspond to different nuclei. Consider four reactions:
where

$\varepsilon$ is the energy released? In which reactions is

$\varepsilon$ positive?

0%
$A + B \rightarrow C + \varepsilon$
0%
$\mathrm { C } \rightarrow \mathrm { A } + \mathrm { B } + \varepsilon$
0%
$D + E \rightarrow F + \varepsilon$
0%
$\mathrm { F } \rightarrow \mathrm { D } + \mathrm { E } + \varepsilon$

In a hydrogen atom, the binding energy of the electron in the nth state is

$E _ { n }$ , then the frequency of revoulation the electron in the nth orbit is

0%
$2 \mathrm { E } _ { \mathrm { n } } / \mathrm { nh }$
0%
$2 \mathrm { E } _ { \mathrm { n } } \mathrm { n }$
0%
$\mathrm { E } _ { \mathrm { n } } / \mathrm { nh }$
0%
$\mathrm { E } _ { \mathrm { n } } \mathrm { n } / \mathrm { h }$

The binding energy per nucleon for

$O^{16}$ and

$O^{17}$ are respectively 7.99 MeV and 7.95 MeV. The energy required to remove one neutron from the nucleus of

$O^{17}$ expressed in MeV is

0%
7.31
0%
4.23
0%
7.13
0%
15.94

If radius of the

$_{ 13 }^{ 27 }{ Al }$ nucleus is estimated to be

$3.6$ fermi, then the radius of

$^{ 125 }{ Te }$ nucleus be nearly

0%
$6$ fermi
0%
$8$ fermi
0%
$4$ fermi
0%
$5$ fermi

If the nucleus is seen as a cubical box of length

$10^{-14}\ m$ , then compute the minimum energy of a nucleon confined to the nucleus. Mass of nucleon=

$1.6\times 10^{-27}\ Kg$

0%
$10\ MeV$
0%
$10\ KeV$
0%
$6\ MeV$
0%
$6\ KeV$

Ratio of binding energy of a satellite revolving few metre above the earth's surface and same satellite rest on earth is

0%
1 : 2
0%
2 : 1
0%
1 : 4
0%
1 : 1

Calculate the binding energy per nucleon of

$_ { 17 } ^ { 35 } \mathrm { C } \ell$ nucleus. Given that mass of

$_ { 17 } ^ { 35 } \mathrm { C } \ell$ nucleus

$= 34.98000 \mathrm { u } ,$ mass of proton

$= 1.007825 \mathrm { u } ,$ mass of neutron

$= 1.008665 \mathrm { u }$ and 1

$\mathrm { u }$ is equivalent to 931

$\mathrm { Mev. }$

0%
6.2 $\mathrm { Mev }$
0%
7.2 $\mathrm { Mev }$
0%
4.2 $\mathrm { Mev }$
0%
8.2 $\mathrm { Mev }$

Which of the following elements is non- radioactive?

0%
plutonium
0%
Zirconium
0%
Thorium
0%
Uranium

An X-ray pulse of wavelength

$4.9 A^{o}$ is sent through a section of Wilson cloud chamber containing a supersaturated gas, and tracks of photoelectron ejected from the gaseous atoms are observed. Two groups of tracks of lengths

$1.40 cm$ and

$2.02 cm$ are noted. If the range energy relation for cloud chamber is given by

$R = aE$ with

$\alpha = \frac { 1 \mathrm { cm } } { k e V }$ . Obtain the binding energies of the two levels from which electrons are emitted. Given

$h = 6.63 \times 10 ^ { - 34 } \mathrm { J } - \mathrm { s } , \mathrm { e } = 1.6 \times 10 ^ { - 19 } \mathrm { J }$

0%
$\phi_{1} = 1.14 KeV$ $\phi_{2}=0.52 KeV$
0%
$\phi_{1} = 1.14 KeV$ $\phi_{2}=0.75 KeV$
0%
$\phi_{1} = 0.52 KeV$ $\phi_{2}=0.75 KeV$
0%
$\phi_{1} = 1.14 KeV$ $\phi_{2}=0.52 KeV$

The binding energy of deutron is

$2.23MeV$ . What should be its mass defect in the units of

$amu$ or

$u$ ?

0%
$0.0012$
0%
$0.0024$
0%
$0.0242$
0%
$0.2422$

A radioactive element

$X$ converts into another stable element

$Y .$ Half-life of

$X$ is 2

$h$ . Initially,only

$X$ is present. After time

$t$ , the ratio of atoms of

$X$ and

$Y$ is found to be

$1 : 4 .$ Then

$t$ inhours is

0%
$2$
0%
$4$
0%
between $4$ and $6$
0%
$6$

In an

$\alpha$ -decay the Kinetic energy of

$\alpha$ particle is 48 MeV and Q-value of the reaction is 50 MeV. The mass number of the mother nucleus is : (Assume that daughter nucleus is in ground state)

0%
96
0%
100
0%
104
0%
none of these

From the following equations pick out the possible nuclear fusion reactions

0%
$_{ 6 }{ C }^{ 13 }+_{ 1 }{ H }^{ 1 }\rightarrow _{ 6 }{ C }^{ 14 }+1.3\quad MeV$
0%
$_{ 6 }{ C }^{ 12 }+_{ 1 }{ H }^{ 1 }\rightarrow _{ 7 }{ C }^{ 14 }+2\quad MeV$
0%
$_{ 7}{ C }^{ 14 }+_{ 1 }{ H }^{ 1 }\rightarrow _{ 8 }{ C }^{ 15 }+7.3\quad MeV$
0%
$_{ 92 }{ U }^{ 235 }+_{ 0 }{ n }^{ 1 }\rightarrow _{ 54 }{ X }^{ 140 }+36{ Sr }^{ 94 }+_{ 0 }{ n }^{ 1 }+_{ 0 }{ n }^{ 1 }+y+200\quad MeV$

${{\text{Bi}}^{{\text{210}}}}$ has half life of 5 days. The time taken for

$\dfrac{7}{8}$ th of a sample to decay is___ days

0%
15
0%
3.4
0%
10
0%
20

The binding energy per nucleon for a deuterium is 1.115 MeV. Mass deflect for this nucleus is about -

0%
2.23 u
0%
0.0024
0%
0.027 u
0%
0.0012 u

Atomic weight of boron is 10.81 and it has two isotopses

$_5B^{10} and _5B^{11}$ . then the ratio of

$_5B^{10} : _5B^{11}$ in nature would be

0%
15 : 16
0%
10: 11
0%
19 : 81
0%
81 ; 19

The binding energy per nucleon for a $_6{C^{12}}$ the nucleus is

( nuclear mass of $_6{C^{12}}$ =12.00000 a.m.u.

Mass of hydrogen nucleus = 1.007825 a.m.u.

Mass of neutron =1.00 8665 a.m.u)

0%
2.675 MeV
0%
7.675 MeV
0%
0 MeV
0%
3.675 MeV

Let

${ m }_{ p }$ be the mass of a proton,

${ m }_{ n }$ the mas of a neutron,

${ M }_{ 1 }$ the mass of a

$^{20}_{10}Ne$ nucleus and

${ M }_{ 2 }$ the mass of a

$^{ 40 }_{ 20 }Ca$ nucleus. Then

0%
${ M }_{ 2 }>2{ M_1 }$
0%
${ M }_{ 2 }<2{ M_{ 1 } }$
0%
${ M }_{ 2 }=2{ M }_{ 1 }$
0%
${ M }_{ 1 }>10\left( { m }_{ n }+{ m }_{ p } \right)$

For

${\text{N = }}{{\text{N}}_0}{e^{ - kt}}and\,{t_2} > {t_1},$ the number of nuclei disintegrating between

${t_1}\,and\,{t_2}$ is..............

0%
${{\text{N}}_0}[{e^{ - \lambda t_1}} - {e^{ - \lambda t_2}}]$
0%
${{\text{N}}_0}[{e^{ - \lambda t_2}} - {e^{ - \lambda t_1}}]$
0%
${{\text{N}}_0}[{e^{\lambda t_2}} - {e^{\lambda t_1}}]$
0%
none of the above.

How much energy is released when

$1$ amu of mass is annihilated?

0%
$931.5\,MeV$
0%
$1.49 \times {10^{ - 3}}J$
0%
$4.138 \times {10^{ - 17}}\,kWh$
0%
All of these

$M(A,Z)$ ,

$M_p$ and

$M_n$ denote the masses of the nucleus

$^A_ZX ,$ proton and neutron respectively in units of

$u$

$( 1u = 931.5 MeV /C^2 )$ and BE represents its bonding energy in

$MeV$ , then

0%
$M(A, Z) = ZM_p + (A-Z)M_n + BE/C^2$
0%
$M(A, Z) = ZM_p + (A-Z)M_n - BE/C^2$
0%
$M(A, Z) = ZM_p + (A-Z)M_n + BE$
0%
$M(A, Z) = ZM_p + (A-Z)M_n - BE$

If the mass of proton =

$1.008 amu$ and mass of neutron =

$1.009 a.m.u.$ , then binding energy per nucleon for

$_{4}Be^{9}$ (mass =

$9.012amu$ ) would be:-

0%
$0.065 MeV$
0%
$60.44MeV$
0%
$67.2MeV$
0%
$6.72MeV$

Explanation

Nuclear Binding energy can be defined of the minimum amount of energy required to break down or disassemble an atom's nuclear into the subatomic

particles that constitutes it

Given: mass of proton

$=1.008$ a.m.u mass of neutron

$=1.009 \mathrm{a.m.u}$

we know,

mass defect is

$\Delta m$

$\Delta m=4 m_{p}+(9-4) m_{n}-m_{B e} \\$

$\Delta m=(4 \times 1.008+5 \times 1.009-9.012) a \cdot m \cdot u \\$

$\Delta m=0.065 a \cdot m \cdot u$

Binding energy will be,

$\cdot E=\Delta m \times 931 \cdot 5 \mathrm{MeV}$

$B \cdot E=0.065 \times 931.5 \\$

$B \cdot E=60.5475 \mathrm{MeV}$

Binding energy per nucleon will be

$\frac{B \cot E}{\text { nucleon }} \Rightarrow \frac{60 \cot 5475}{9}$

Binding energy per nucleon is

$6.7275 \mathrm{MeV}$

$option [D] 6.72 Mev$

The mass of proton is

$1.0073 u$ and that of neutron is

$1.0087 u$ (

${ u } =$ atomic mass unit). The binding energy of

$_ { 2 } H e ^ { 4 }$ is (Given, mass of helium nucleus

$= 4.0015{ u }$ )

0%
$28.4{ MeV }$
0%
$62.4{ MeV }$
0%
$42.4{ MeV }$
0%
$10.2{ MeV }$

The decay constant of the end product if a radioactive series is

0%
zero
0%
infinite
0%
finite ( non zero)
0%
depends on the end product

Free

$^{238}U$ nuclei kept in a train emit alpha particles. When the train is stationary and a uranium nucleus decays, a passenger measures that the separation between the alpha particle and the recoiling nucleus becomes x in time t after the decay. If a decay takes place when the train is moving at a uniform speed v, the distance between the alpha particle and the recoiling nucleus at a time t after the decay, as measured by the passenger will be

0%
x + vt
0%
x- vt
0%
x
0%
depends on the direction of the train.

If ionization potential of Hydrogen atom is 13.6 V then what is ionization potential of He atom ?

0%
27.6 V
0%
13.6 V
0%
54.2 V
0%
None of these

Explanation

We know

Lionization energy is directly proportional to square of atomic no. ie

$\left(I \cdot E \propto^{2}\right)$ , so

$\frac{(I E)_{H e}}{(I E)_{H}}$ =

$\frac{\left(z^{2}\right)_{H e}}{\left(z^{2}\right)^{2} H}$

$(I E)_{H e}$ =

$\frac{(2)^{2}}{(1)^{2}} \times 13.6$

$(I \cdot E)_{H e}$ =

$54 \cdot 2 \mathrm{~V} \\$

${[c] 54 \cdot 2 \mathrm{v}}$

Using the following data mass of hydrogen atom

$=1.00783u$ mass of neutron

$1.00867u$ mass of nitrogen atom

$(_{ 7 }N^{14})$ =

$14.00307u$ . The calculated value of the binding energy of the nucleus of the nitrogen atom

$(_{ 7 }N^{ \ 14\ })$ is close

0%
$56MeV$
0%
$98MeV$
0%
$104MeV$
0%
$112MeV$

Explanation

Nuclear Binding energy can be defined es the minimum

amount of energy required to break down or

disassemble an particles that constitutes it.

m=14.00307 u

A nucleus of nitrogen

$7^{N^{14}}$ contains 7 protons and

7 neutrons

Hence, the mass defect of this nuclers will be,

$\Delta m=7 m_{H}+7 m_{n}-m$

$m_{n}=1.008665 \mathrm{~m}$

$\therefore \Delta m=7 \times 1.007825+7 \times 1.008665-14.00307$

$\Delta m$ =

$7.054775+7.06055-14.00307 \\$

$\Delta m$ =

$0.11236 \mathrm{~m}$

we know,

$1 u=931.5 \text { Mev } 1 c^{2}$

$\therefore \quad \Delta m=0.11236 \times 931.5 \mathrm{MeV} / \mathrm{c}^{2}$

Hence, the binding energy of the nucleus is given as:

$E_{b} \Delta m c^{2}$ where,

$c=$ speed of light

$\therefore E_{b}=0.112 .36 \times 931.5\left(\frac{M e V}{c^{2}}\right) ^{2}$

[C]

$E_{b}=104 \mathrm{MeV}$

Which element is the end product of each natural radioactive series?

0%
Sn
0%
Bi
0%
Pb
0%
C

Explanation

Three of the sets, the thorium series, uranium-series, and actinium series, called natural or classical series.

The thorium series begins with thorium-232 and ends with the stable nuclide lead-208.

The uranium-series begins with uranium-238 and ends with lead-206.

The actinium series, named for its first-discovered member, actinium-227, begins with uranium-235 and ends with lead-207.

So, the end product of each natural radioactive series is lead.

The answer is option

$C$ .

Choose the element which is not radioactive.

0%
Cm
0%
No
0%
Mo
0%
Md

Explanation

Radioactive elements are unstable isotopes that release subatomic particles or energy as they decay.

Curium(Cm), Nobelium (No), Mendelevium(Md) are transuranic radioactive chemical elements. But molybdenum (Mo) is not a radioactive element.

Hence, option

$C$ is correct.

Nd (Z = 60) is a member of group -3 in the periodic table. An isotope of it is

$\beta$ -active. The daughter nuclei will be a member of:

0%
Group - 3
0%
Group - 4
0%
Group - 1
0%
Group - 2

Radioactivity was discovered by

0%
Henry Becquerel
0%
Rutherford
0%
J. J. Thomson
0%
Madam Curie

Explanation

It was Henri Becquerel that discovered radioactivity, but it was Marie Curie who coined the term.

Hence, option

$A$ is correct.

Tritium undergoes radioactive decay giving:

0%
$\alpha$ - particles
0%
$\beta$ - particles
0%
Neutrons
0%
None of these

Explanation

Tritium

$(_1H^3 \rightarrow _2He^3 + _{-1}e^0)$ is a

$\beta$ - emitter.

The phenomenon of radioactivity arises from the

0%
Binary fission
0%
Nuclear fusion
0%
Stable nuclei
0%
Decay of unstable nuclei

If radium and chlorine combine to from radium chloride the compound is:

0%
No longer radioactive
0%
Twice as radioactive as radium
0%
Half as radioactive as radium
0%
As radioactive as radium

Explanation

Radioactivity is a nuclear property.

The formation of a compound from a radioactive element does not involve any change in the structure of nucleus . Thus, radium chloride will be as radioactive as radium itself.

Hence, option

$D$ is correct.

Which of the following statements about the radioactivity of an element is incorrect?

0%
It is a nuclear property
0%
It does not involve any rearrangement of electrons
0%
Its rate is affected by change in temperature and/or pressure
0%
It remains unaffected by the presence of other element or elements chemically combined with it

When a sample of solid lithium is placed in a flask of hydrogen gas then following reaction happened

$_1^1H+_3Li^7 \rightarrow He^4+_2He^4$
This statement is

0%
True
0%
False
0%
May be true at a particular presure
0%
None of these

Explanation

The given reaction is a nuclear reaction, which can take place only if a proton (a hydrogen nucleus) comes into contact with a lithium nucleus. If the hydrogen is in the atomic from, the interaction between it's electron cloud and the electron cloud of a lithium atom keeps the two nuclei from getting close to each other. Each if isolated protons are used, they must be fixed at the

$Li$ atom with enough kinetic energy to over come the electric repulsion between the proton and

$Li$ atom.

Radioactivity is affected by:

0%
temperature
0%
pressure
0%
electric and magnetic field
0%
none of these

Explanation

$\begin{array}{l} \text { Temperature, pressure, electric and magnetic field } \\ \text { does not affect Radioactivity. } \\ \text { The correct is option D. } \end{array}$

One or more answers is/are correct.
In electron capture (radioactive process)

0%
a neutron is formed
0%
a proton is consumed
0%
$\gamma$ -ray emission takes place
0%
X-ray emission takes place

Explanation

$_{1}^{1}P + _{-1}e^{0} \rightarrow _{0}^{1}n + X-rays$

Hence option (A) and (B) is correct.

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 13 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 13 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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