MCQ Questions for Class 12 Medical Physics Nuclei Quiz 10

Among

$\alpha$ -decay and

$\beta$ -decay which cause a change of element?

0%
Both $\alpha$ -decay and $\beta$ -decay
0%
Only $\alpha$ -decay
0%
Only $\beta$ -decay
0%
None of them.

The nucleus of mass

$M + \Delta m$ is at rest and decays into two daughter nuclei of equal mass

$\dfrac { M } { 2 }$ each. Speed of light is

$c.$ This binding energy per nucleon for the parent nucleus is

$E _ { 1 }$ and that for the daughter nuclei is

$E _ { 2 }.$ Then :

0%
$E _ { 1 } = 2 E _ { 2 }$
0%
$E _ { 1 } > E _ { 2 }$
0%
$E _ { 2 } > E _ { 1 }$
0%
$E _ { 2 } = 2 E _ { 1 }$

Write the S.I. unit of activity.

0%
$Becquerel$
0%
$Henry$
0%
$Ohm$
0%
$Mendel$

Explanation

$S.I.$ Unit of activity : It is the count rate of radioactivity.

It's

$S.I$ Unit is decay per second

$=1\ Bq$

where

$Bq=Becquerel (\because 1\ Bq=1\ decay\ per second)$

The wavelength of

$K _ { \alpha }$ line for an element of atomic number

$43$ is

$\lambda.$ Then the wavelength of

$K _ { \alpha }$ line for an element of atomic number

$29$ is

0%
$\dfrac { 43 } { 29 } \lambda$
0%
$\dfrac { 42 } { 28 } \lambda$
0%
$\dfrac { 9 } { 4 } \lambda$
0%
$\dfrac { 4 } { 9 } \lambda$

Explanation

As we know,

$\lambda \propto \dfrac{1}{(Z-1)^2} \Rightarrow \dfrac{\lambda_2}{\lambda_1}=\left(\dfrac{Z_1-1}{Z_2-1}\right)^2$

$\Rightarrow \dfrac{\lambda_2}{\lambda}=\left(\dfrac{43-1}{29-1}\right)^2=\left(\dfrac{42}{28}\right)^2 \Rightarrow \lambda_2=\dfrac{9}{4} \lambda$

A scientist carries out an experiment using a sealed source which emits

$\beta$ -particles. The range of the

$\beta$ - particles in the air is about

$30cm$ .
Which precaution is the most effective to protect the scientist from the radiation?

0%
handling the source with long tongs
0%
keeping the temperature of the source low
0%
opening all windows in the laboratory
0%
washing his hands before leaving the laboratory

Explanation

Handling the source with long tongs will prevent him from the exposure of

$\beta$ - particles. Since the range of

$\beta$ -particle is only

$30cm$ so, long tongs can easily be a safer and effective precaution.

Protons of energy

$2$ eV and

$2.5$ eV successively illuminate a metal whose work function is

$0.5$ eV. The ratio of maximum speed of emitted electron is _______.

0%
$\sqrt{3}:2$
0%
$2:1$
0%
$1:2$
0%
$2:\sqrt{3}$

Explanation

$K.E_{max}=hv-\phi$

$\dfrac{1}{2}mv^2=hv-\phi$

$\dfrac{v^2_1}{v^2_2}=\dfrac{2-0.5}{2.5-0.5}=\dfrac{1.5}{2}=\dfrac{3}{4}$

$\dfrac{v_1}{v_2}=\dfrac{\sqrt{3}}{2}$ .

A sample containing same number of two nuclei A and B start decaying. The decay constant of A and B are

$10 \lambda$ and

$\lambda$ . The time after which

$\dfrac{N_A}{N_B}$ becomes

$\dfrac{1}{e}$ is

0%
$\dfrac{1}{9 \lambda}$
0%
$\dfrac{1}{18 \lambda}$
0%
$\dfrac{2}{9 \lambda}$
0%
$\dfrac{3}{19 \lambda}$

Explanation

$\dfrac{N_A}{N_B} = \dfrac{N_0 e^{-10 \lambda t}}{N_0 e^{-\lambda t}} = \dfrac{1}{e}$

$\Rightarrow e^{-9 \lambda t} = e^{-1}$

$\Rightarrow 9 \lambda t = 1$

$\Rightarrow t = \dfrac{1}{9 \lambda}$

The notation for an isotope of sodium

$^{23}_{11}Na$ . Which row gives the composition of a neutral atom of this isotope of sodium?

	number of protons	number of neutrons	number of electrons
A	$11$	$12$	$11$
B	$11$	$12$	$12$
C	$11$	$23$	$11$
D	$12$	$11$	$12$

0%
A
0%
B
0%
C
0%
D

Explanation

$^{23}_{11}Na$

Atomic No. ,

$A= 11$

Mass No. ,

$M= 23$

Number of protons=

$A= 11$

As it its not ionized
Number of electrons

$=A=11$

Number of neutrons,

$=M-A=23-11=12$

For mass defect of

$0.3\%$ , the binding energy of

$1kg$ material is :

0%
$2.7\times 10^{14}erg$
0%
$2.7\times 10^{14}J$
0%
$2.7\times 10^{-14}erg$
0%
$2.7\times 10^{-14}J$

Explanation

$\Delta m=0.3 \% \text { of } 1 \mathrm{~kg} \\$

$\Delta m=\frac{0.3}{100} \\$

$\Delta m=3 \times 10^{-3} \mathrm{~kg} \\$

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

$E=3 \times 10^{-3} \times\left(3 \times 10^{8}\right)^{2} \\$

$E=3 \times 9 \times 10^{-3} \times 10^{16} \mathrm{~J} \\$

${[B] E=2.7 \times 10^{14} \mathrm{~J}}$

Molecular mass of dry air is ________________.

0%
less than moist air
0%
greater than moist air
0%
equal to moist air
0%
may be greater or less than moist air

Explanation

The molecular mass of dry ice is greater than moist air.

As dry air consists of nitrogen and oxygen while moist air contains water vapour which less mass than that of oxygen and nitrogen making the mass of moist air lower.

$M.M$ of dry air

$\simeq 28.9\ gm$

$M.M$ of moist air

$< 28.9\ gm$

Which diagram represents a nucleus of

$^3_1H$ ?

Explanation

Here

$=$ a neutron

$=$ a proton.

$^3_1 H$

As its Atomic No. is

$1$ . So the number of protons is

$1$

And its Mass no is

$3$

No. of neutrons = Atomic No. - Mass No.

$=3-1=2$

So follwing diagram is correct structure of

$^3_1H$

1333383_1645241_ans_514e05d7b81a4b44b6e1acc8e013ee32.png

When the atomic number

$A$ of a nucleus increases :

0%
initially the neutron-proton ratio is constant
0%
initially the neutron-proton per nucleon increases and then decreases
0%
initially the binding energy per nucleon increases and then decreases
0%
the binding energy per nucleon increases when neutron proton ratio increases

The nuclear radius is given by

$R = r_0A^{1/3}$ , where

$r_0$ is constant and A is the atomic mass number.
Then :

0%
The nuclear mass density of $U^{238}$ is twice that of $Sn^{119}$ .
0%
The nuclear mass density of $U^{238}$ is thrice that of $Sn^{119}$
0%
The nuclear mass density of $U^{238}$ is the same as that of $Sn^{119}$
0%
The nuclear mass density of $U^{238}$ is half that of $Sn^{119}$

Explanation

$|C|$

$R=r_0A^{1/3}$

$R^3=r^3_0A$

$Density=\dfrac{Mass}{Volume}=\dfrac{Am_p}{\dfrac{4}{3}\pi R^3}=\dfrac{Am_p}{\dfrac{4}{3}\pi r^3_0A}=\dfrac{3m_p}{4\pi r^3_0}$

Hence mass density does not depend on mass (A) or atomic number (z)

The atomic mass of

$B^{10}$ is

$10.811\ amu$ . The binding energy of

$B^{10}$ nucleus is [ Given: The mass of electron is

$0.0005498\ amu$ , the mass of proton is

$m_p=1.007276\ amu$ and the mass of neutron is

$m_n=1.008665\ amu$ ]:

0%
$-678.272\ MeV$
0%
$678.932\ MeV$
0%
$378.932\ MeV$
0%
$none\ of\ these$

Radioactivity can be effected by :

0%
temperature
0%
pressure
0%
radiation
0%
all of these

A binding energies per nucleon for a deuteron and an

$\alpha-$ particle are

$x_1$ and

$x_2$ respectively. What will be the energy

$Q$ released in the reaction?

$_{1}H^{2}+_{1}H^{2}\longrightarrow _{2}He^{4}+Q$

0%
$4(x_1+x_2)$
0%
$4(x_2-x_1)$
0%
$2(x_1+x_2)$
0%
$2(x_2-x_1)$

Explanation

The correct option is B.

Given,

A binding energies per nucleon for deuteron and an

$\alpha$ - particle are

$x_1$ and

$x_2$ respectively.

The provided reaction is

${H_{1}}^{2}+{H_{1}}^{2}\rightarrow {H_{2}}^{4}+Q$

We are required to find out the energy released in the reaction i..e Q value.

The number of deuterons is

$2$ .

The number of nucleons for

$2$ deuteron is

$4$ .

The number of nucleons for

$2$ helium is

$4$ .

The energy for

${H_{1}}^{2}$ is

$4x_1$ .

The energy for

${H_{2}}^{4}$ is

$4x_2$

Therefore, the energy released is calculated as

$Q$ =

$4\left ( x_2-x_1 \right )$ .

Hence, the required answer is

$Q$ =

$4\left ( x_2-x_1 \right )$ .

The binding energy of

$Na^{23}$ is [ Given : Atomic mass of

$22.9898\ amu$ and that of

$^1 H_1$ is

$1.00783\ amu$ ]:

0%
$931\ MeV$
0%
$186.54\ MeV$
0%
$5.38\ MeV$
0%
$none\ of\ these$

The critical mass of fissionable uranium

$235$ can be reduced by:

0%
surrounding it by neutron absorbing material
0%
surrounding it by neutron reflecting material
0%
heating the material
0%
adding impurities

As comparee to

$^{12}C$ atom ,

$^{14}C$ atom has

0%
two extra protons and two extra electrons
0%
two extra protons but no extra electrons
0%
two extra neutrons and no extra electron
0%
two extra neutron and two extra electrons

As the mass number A increases, the binding energy per nucleon in a nucleus

0%
increases
0%
decreases
0%
remains the same
0%
varies in a way that depends on the actual value of A

For nuclei with A >100,

0%
The binding energy of the nucleus decreases on an average as A increases
0%
The binding energy per nucleon decreases on an average as A increases
0%
If the nucleus breaks into two roughly equal parts, energy is released
0%
If two nuclei fuse to form a bigger nucleus, energy is released.

Isotopes, which undergo spontaneous fission are found in

$n-p$ graph :

0%
above the belt of stability
0%
below the belt of stability
0%
above or below the belt of stability
0%
in the belt of stability

When number of nucleons in nuclei increases, the binding energy per nucleon:

0%
increases continuously with mass number
0%
decreases continuously with mass number
0%
remain constant with mass number
0%
first increases and then decreases with increase of mass number

$\alpha,\beta$ and

$\gamma$ radiations come out of a rodiactive substance

0%
when it is heated
0%
when put in atomic reactor
0%
spontaneously
0%
under pressure

In nuclear reaction

$_{2}He^{4}+\ _{Z}X^{A}\rightarrow Z+\ _{2}\gamma^{A+3}+\ _{Z}M^{A}$
where

$M$ denotes

0%
electron
0%
positron
0%
proton
0%
neutron

A proton and an alpha particle having same momentum enter a magnetic field at right angles to it. If

$r_1$ and

$r_2$ be their radii respectively then value of

$r_1 /r_2$ is :

0%
$1$
0%
$2$
0%
$1/2$
0%
$1/4$

Group displacement law states that the emission of

$\alpha$ or

$\beta$ particles results in the daughter element occupying a position, in the periodic table, either to the left or right of that of the parent element.

Which one of the following alternatives gives the correct position of the daughter element?

0%
On emission of $\alpha$ particle - 2 groups to the right;
On emission of $\beta$ particle- 1 group to the right
0%
On emission of $\alpha$ particle - 2 groups to the right;
On emission of $\beta$ particle- 1 group to the left
0%
On emission of $\alpha$ particle - 2 groups to the left;
On emission of $\beta$ particle- 1 group to the left
0%
On emission of $\alpha$ particle - 2 groups to the left;
On emission of $\beta$ particle- 1 group to the right

Explanation

Alpha emission decreases the atomic number by two. So, on the emission of

$\alpha$ -particle daughter element shift 2 group to the left.

Beta emission increases the atomic number by one. So, on the emission of

$\beta$ -particle daughter element shift 1 group to the right.

Hence, option

$D$ is correct.

The

$_{88}Ra^{226}$ is:

0%
n-mesons
0%
u-mesons
0%
Radioactive
0%
Non-radioactive

Explanation

$_{88}Ra^{226}$ is radioactive because

$\dfrac{n}{p}$ ratio for it is 1.56 which is greater than 1.5.

Nuclear reactivity of

$Na$ and

$Na^+$ is same because both have:

0%
Same electron and proton
0%
Same proton and same neutron
0%
Different electron and proton
0%
Different proton and neutron

Explanation

Nuclear reactivity depends upon

$\frac{n}{p}$ ratio and

$\frac{n}{p}$ ratio is same for both

$Na$ and $$Na^+$$, so they both have same nuclear reactivity.

Hence, option

$B$ is correct.

Which one of the following nuclear reaction is correct?

0%
$_6C^{13} + _1H^1 \rightarrow _7N^{13} + \beta^- + v^-$
0%
$_{11}Ba^{23} + _1H^1 \rightarrow _{10}Ne^{20} + _2He^4$
0%
$_{13}Al^{23} + _0n^1 \rightarrow _{11}Na^{23} + e^0$
0%
None of these

Explanation

The sum of atomic masses and the sum of atomic numbers on the reactant side is equal to the sum of atomic masses and the sum of atomic numbers on the product side respectively.

In option

$A$ , the mass number on both sides is not equal. In option

$C$ , atomic number on both sides is not equal. Only option

$B$ is a correct reaction.

Hence, option

$B$ is correct.

In the nuclear reaction

$_{92}U^{238} \rightarrow _{82}Pb^{206} + x \ _2He^4 + y \ _{-1} \beta^0$
the value of x and y are respectively ........

0%
8, 6
0%
6, 4
0%
6, 8
0%
8, 10

Explanation

The sum of atomic masses and the sum of atomic numbers on the reactant side is equal to the sum of atomic masses and the sum of atomic numbers on the product side respectively.

Sum of atomic masses of reactants = sum of atomic masses of products

$238\ =\ 206+4x+0$

$4x=32$

$x=8$

Sum of atomic numbers of reactants = sum of the atomic numbers of products

$92=82+2x+y(-1)$

$10=(2\times8)-y$

$y=6$

Hence, the option

$A$ is correct.

Which of the following is radioactive element?

0%
Sulphur
0%
Polonium
0%
Tellurium
0%
Selenium

Radioactivity is due to:

0%
Stable electronic configuration
0%
Unstable electronic configuration
0%
Stable nucleus
0%
Unstable nucleus

Explanation

Radioactive decay (radioactivity) occurs in unstable atomic nuclei that don't have enough binding energy to hold the nucleus together due to an excess of either protons or neutrons.

Hence, option

$D$ is correct.

When a radioactive element emits an electron, the daughter element formed will have:

0%
Mass number one unit less
0%
Atomic number one unit less
0%
Mass number one unit more
0%
Atomic number one unit more

Explanation

A beta particle is an electron emitted by a radioactive nucleus. In beta decay, one of the neutrons in the nucleus suddenly changes into a proton, causing an increase in the atomic number of the parent element.

$_z A^m \rightarrow \ _{z + 1} B^m + \ _{-1} e^0$

Hence, option

$D$ is correct.

In the nuclear reaction

${ _{ 1 }^{ }{ H } }^{ 2 }+{ _{ 1 }^{ }{ H } }^{ 2 }\rightarrow { _{ 2 }^{ }{ He } }^{ 3 }+{ _{ 0 }^{ }{ n } }^{ 1 }$
if the mass of the deuterium atom

$=2.014741a.m.u$ , mass of

${ _{ 2 }^{ }{ He } }^{ 3 }=3.016977a.m.u$ and mass of neutron

$=1.008987a.m.u$ , then the

$Q$ value of the reaction is nearly

0%
$0.00352MeV$
0%
$3.27MeV$
0%
$0.82MeV$
0%
$2.45MeV$

Explanation

$Q=-\Delta mc^2$

$Q=\left( \sum { { M }_{ r } } -\sum { { M }_{ p } } \right) { c }^{ 2 }$

$\sum { { B }_{ r } } =2\times 2.014741a.m.u=4.0294892a.m.u$

$\sum { { B }_{ p } } =(3.016977+1.008987)a.m.u=4.025964a.m.u$

$\sum { { B }_{ r } } -\sum { { B }_{ p } } =0.003518a.m.u\quad$

The decrease in mass appears as equivalent energy

$Q=0.003518\times 931MeV=3.27MeV\quad$

Consider one of fission reactions of

$_{ }^{ 235 }{ U }$ by thermal neutrons

$_{ 92 }^{ 235 }{ U }+n\rightarrow _{ 38 }^{ 94 }{ Sr }+_{ 54 }^{ 140 }{ Xe }+2n$ . The fission fragments are however unstable and they undergo successive

$\beta$ - decay until

$_{ 38 }^{ 94 }{ Sr }$ becomes

$_{ 40 }^{ 94 }{ Zr }$ and

$_{ 54 }^{ 140 }{ Xe }$ becomes

$_{ 58 }^{ 140 }{ Ce }$ . The energy released in this process is
[Given:

$m(_{ }^{ 235 }{ U })=235.439u$ ;

$m(n)=1.00866u$ ;

$m(_{ }^{ 94 }{ Zr }=93.9064u$ ;

$m(_{ }^{ 140 }{ Ce })=139.9055u$ ;

$1u=931MeV$ ]

0%
$156MeV$
0%
$208MeV$
0%
$456MeV$
0%
Cannot be computed

Explanation

$_{92}^{235}U+n\longrightarrow _{38}^{94}Sr+_{54}^{140}Xe+2n$

$_{38}^{94}Sr\longrightarrow _{40}^{94}Zr+2e^-$

$_{54}^{140}Xe\longrightarrow _{58}^{140}Xe+4e^-$

The complete fission reaction is

$_{ 92 }^{ 235 }{ U }+n\rightarrow _{ 40 }^{ 94 }{ Zr }+_{ 58 }^{ 140 }{ Ce }+2n+6{ e }^{ -1 }$

$Q=\Delta mc^2$

$Q=\left[ m\left( _{ }^{ 235 }{ U } \right) -m\left( _{ }^{ 94 }{ Zr } \right) -m\left( _{ }^{ 140 }{ Ce } \right) -m(n) \right] { c }^{ 2 }=208MeV$

$1.00kg$ of

$_{ }^{ 235 }{ U }$ undergoes fission process. If energy released per event is

$200MeV$ , then the total energy released is

0%
$5.12\times {10}^{24}MeV$
0%
$6.02\times {10}^{23}MeV$
0%
$5.12\times {10}^{26}MeV$
0%
$6.02\times {10}^{26}MeV$

Explanation

The number of nuclei in

$1kg$

$_{ }^{ 235 }{ U }$ is

$N=\cfrac { { N }_{ A } }{ 235 } \times \left( 1\times { 10 }^{ 3 } \right) =2.56\times { 10 }^{ 24 }$ nuclei
Total energy released is

$E=N\times 200MeV=5.12\times { 10 }^{ 26 }MeV$

Binding energy per nucleon of

${ _{ 1 }^{ }{ H } }^{ 2 }$ and

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

$1.1MeV$ and

$7.0MeV$ , respectively. Energy released in the process

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

0%
$20.8MeV$
0%
$16.6MeV$
0%
$25.2MeV$
0%
$23.6MeV$

Explanation

Given

$BE_{He}=4\times7.0MeV=28.0MeV$

$BE_H=2\times1.1Mev=2.2MeV$

Energy released in the reaction is equal to the total binding energy of reactants minus the total binding energy of products.

Hence, Energy released would be

$\Delta E=BE_{He}-2\times BE_{H}$

$\Delta E=(28.0-2\times 2.2)MeV$

$\Delta E=23.6MeV$

Calculate the binding energy of a deutron atom, which consists of a proton and a neutron, given that the atomic mass of the deuteron is

$2.014102u$

0%
$0.002388MeV$
0%
$2.014102MeV$
0%
$2.16490MeV$
0%
$2.224MeV$

Explanation

atomic mass

$M(H)$ of hydrogen and nuclear mass

$({M}_{n})$ are

$M(H)=1.007825u$ and

${M}_{n}=1.008665u$
Mass defect

$\Delta m=\left[ M(H)+{ M }_{ n }-M(D) \right] ;M(D)=2.016490u-2.014102u=0.002388u$
As

$1u$ corresponds to

$931.494MeV$ energy, therefore, mass defect corresponds to energy

${ E }_{ b }=0.002388\times 931.5=2.224MeV\quad$

Mark out the correct statement(s)

0%
In alpha decay, the energy released is shared between alpha particle and daughter nucleus in the form of kinetic energy and share of alpha particle is more than that of the daughter
0%
In beta decay, the energy released in the form of kinetic energy of beta particles
0%
In beta minus decay, the energy released is shared between electron and antineutrino
0%
In gamma decay, the energy released is in the form of energy carried by photons termed as gamma rays

Consider the following reaction:

${ _{ }^{ 1 }{ H } }_{ 2 }+{ _{ }^{ 1 }{ H } }_{ 2 }\rightarrow { _{ 1 }^{ }{ He } }^{ 4 }+Q$
If

$m({ _{ 1 }^{ }{ H } }^{ 2 })=2.0141u$ ;

$m({ _{ 2 }^{ }{ He } }^{ 4 })=4.0024u$ , the energy

$Q$ released (in MeV) in this fusion reacion is

0%
12
0%
6
0%
24
0%
48

Explanation

${ _{ 1 }^{ }{ H } }^{ 2 }+{ _{ 1 }^{ }{ H } }^{ 2 }\rightarrow { _{ 1 }^{ }{ He } }^{ 4 }+Q\quad$

$\Delta m=m\left( { _{ 2 }^{ }{ He } }^{ 4 } \right) -2m\left( { _{ 1 }^{ }{ H } }^{ 2 } \right) =-0.0258u$

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

$=\left( 0.0258 \right) \left( 931.5 \right) MeV$

$\approx 24MeV$

If the mass of isotope

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

$7.016005\,u$ and masses of H atom and a neutron are

$1.007825 \,u$ and

$1.008665 \,u$ respectively. The binding of Li nucleus is:

0%
$5.6 \,MeV$
0%
$8.8\,MeV$
0%
$0.42 \,MeV$
0%
$39.2 \,MeV$

Explanation

$\Delta m=[3 \times 1.00725\ u+4 \times 1.008665\ u ]- 7.016005\ u$

$\Delta m=[3.023475\ u+4.03466\ u ]- 7.016005\ u$

$\Delta m=[7.058135\ u - 7.016005\ u$

$\Delta m =0.042134\ u$

$E_B= \Delta m c^2$

$E_B=0.04213 \times 931.5 \dfrac{MeV}{c^2}c^2$

$E_B=39.24\ MeV$

Which of the following graphs might represent the relationship between atomic number (i.e, "atomic weight") and the total binding energy of the nucleus, for nuclei heavier than

$_{38}^{94}Sr$ ?

Choose the correct statements from the following:

0%
Like other light nuclei, the ${ _{ 2 }^{ }{ He } }^{ 4 }$ nuclei also have a low value of the binding energy per nucleon
0%
The binding energy per nucleon decreases for nuclei with small as well as large atomic number
0%
The energy required to remove one neutron from ${ _{ 3 }^{ }{ Li } }^{ 7 }$ to transform it into the isotope ${ _{ 3 }^{ }{ Li } }^{ 6 }$ is $5.6MeV$ , which is the same as the binding energy per nucleon of ${ _{ 3 }^{ }{ Li } }^{ 6 }$
0%
When two deuterium nuclei fuse together, they give rise to a tritium nucleus accompanied by a release of energy

Explanation

Statement (a) is incorrect. The

${ _{ 2 }^{ }{ He } }^{ 4 }$ nucleus (or the

$\alpha$ -particle) is exceptionally stable and has a much higher value of BE per nucleon than that for most other light nuclei.

Statement (b) is correct but the reason of decrease in the BE per nucleon for nuclei with large

$A$ is that with an increase in the number of protons, the coulomb repulsion increases. On the other hand, the decrease in the BE per nucleon for nuclei with small

$A$ is due to a surface effect: the nucleons at the surface being less strongly bound than those in the interior.

Statement (c) is also correct. The energy required to remove one neutron (ie one nucleon) is the same as the binding energy per nucleon for a given isotope.

Statement (d) is incorrect. To ensure both charge and mass number conservation, a proton must be produced as a by-product of the reaction:

${ _{ 1 }^{ }{ D } }^{ 2 }+{ _{ 1 }^{ }{ D } }^{ 2 }\rightarrow { _{ 1 }^{ }{ T } }^{ 3 }+{ _{ 1 }^{ }{ P } }^{ 1 }+Q$

Binding energy per nucleon for

${C}^{12}$ is

$7.68MeV$ and for

${C}^{13}$ is

$7.74MeV$ . The energy required to remove a neutron from

${C}^{13}$ is

0%
$5.49MeV$
0%
$8.46MeV$
0%
$49.45MeV$ $
0%
$15.49MeV$

Explanation

$C^{13}\longrightarrow C^{12}+n$

Binding energy of

$C^{12}=12\times7.68MeV=92.16MeV$

Binding energy of

$C^{13}=13\times7.74MeV=100.62MeV$

Energy required to remove a neutron

$=B_{C^{13}}-B_{C^{12}}$

$E=100.62MeV-92.16MeV$

$E=8.46MeV$

When a nucleus in an atom undergoes radioactive decay, the electronic energy levels of the atom

0%
do not change for any type of radioactivity
0%
change for $\alpha$ and $\beta$ radioactivity but not for $\gamma$ - radioactivity
0%
change for $\alpha$ - radioactivity but not for others
0%
change for $\beta$ - radioactivity but not for others

Explanation

$\beta$ particles carries one unit of negative charge, and

$\alpha$ particle carries

$2$ units of positive charge, and Y-particle carries no charge. So the electronic energy level of the atom changes in emission of

$\alpha$ and

$\beta$ particle, but not in

$Y$ decay.

Heavy stable nucleus have more neutrons than protons. This is because of the fact that

0%
neutrons are heavier than protons
0%
electrostatic force between protons are repulsive
0%
neutrons decay into protons through beta decay
0%
nuclear force between neutrons are weaker than that between protons

Fusion processes, like combining two deuterons to form a

$He$ nucleus are impossible at ordinary temperatures and pressure. The reasons for this can be traced to the fact:

0%
nuclear forces have short range
0%
nuclei are positively charged
0%
the original nuclei must be completely ionized before fission can take place
0%
the original nuclei must first break up before combining with each other

Explanation

Two deuteron can combine to form

$He$ atom when their nuclei come close to nuclear range where electrostatic repulsive force between positively charged deuterons does not act. The electrostatic force increases very high on decreasing their distance.

$\left( \because F\propto \cfrac { 1 }{ { r }^{ 2 } } \right)$
To overcome this electrostatic repulsive force nuclei need very high temperature and pressure. Hence to combine two nuclei, they must reach closer of the range of where nuclear force acts and electrostatic repulsive force does not act verifies the answer (a) and (b)

Mass is measure of the amount of matter.

0%
True
0%
False

The particle used in nuclear fission for bombardment is:

0%
Alpha particle
0%
Proton
0%
Beta particle
0%
Neutron

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 10 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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