MCQ Questions for Class 12 Medical Physics Nuclei Quiz 5

Consider

$_{13}^{25}\textrm{Al}$

$\rightarrow$

$_{12}^{25}\textrm{Mg}$ +

$_{+1}^{0}\textrm{e}$ +

$v$ .

$m_{Al}$ =

$24.990423 \mu$ ;

$m_{mg}$ =

$24.485839 \mu$ .
Find the Q value of reaction

0%
$3.3$ $Mev$
0%
$1.3$ $Mev$
0%
$2.3$ $MeV$
0%
$5.3$ $MeV$

Explanation

$Q$

$=[24.990432 \mu-24.98583 \mu-2m_{e}]c^{2}$

$=0.004593(931.5) - 1.102$

$=3.3$

$MeV$

In the fusion process there are

0%
isotopes of hydrogen
0%
isotopes of helium
0%
isotopes of carbon
0%
isotopes

Which of the following is a wrong description of binding energy of a nucleus?

0%
It is the energy required to break a nucleus into its constituent nucleons
0%
It is the energy made available when free nucleons combine to form a nucleus
0%
It is the sum of the rest mass energies of its nucleons minus the rest mass energy of nucleus
0%
It is the sum of the kinetic energy of all the nucelons in the nucleus

The fusion of two nuclide will require a temp of order of

0%
$10^{7}$ $K$
0%
$10^{6}$ $K$
0%
$10^{8}$ $K$
0%
$10^{9}$ $K$

The necessary condition for nuclear fusion is

0%
high temperature and high pressure
0%
low temperature and low pressure
0%
high temperature and low pressure
0%
low temperature and high pressure

Explanation

High temperature - The high temperature gives the hydrogen atoms enough energy to overcome the electrical repulsion between the protons.

High pressure - Pressure squeezes the hydrogen atoms together. They must be within

$1\times10^{-15}$ meters of each other to fuse.

For the fast chain reaction, the size of

$U^{235}$ block, as compared to its critical size, must be

0%
greater
0%
smaller
0%
same
0%
anything

The critical mass of fissionable material is

0%
$75$ $kg$
0%
$1$ $kg$
0%
$20$ $kg$
0%
$10$ $kg$

The value of A in the following reaction is

$_{4}\textrm{Be}^{9}$ +

$_{2}\textrm{He}^{4}$ =

$_{6}\textrm{C}^{A}$ +

$_{0}\textrm{n}^{1}$

0%
$14$
0%
$10$
0%
$12$
0%
$16$

Explanation

The reaction producing neutrons by bombarding beryllium with alpha particles is given as

$_{4}^{9}\textrm{Be} + _{2}^{4}\textrm{He} \rightarrow _{6}^{A}\textrm{C} + _{0}^{1}\textrm{n}$
Since, a neutron has atomic number of zero (it has no positive charge) but has a mass number of one (the same mass as the proton). Hence, the total mass (that is, the total number of nuclear particles or nucleons = 13) entering into the reaction must be the same as nucleons leaving the reaction.
Therefore

$A = 12$ .

$_{4}^{9}\textrm{Be} + _{2}^{4}\textrm{He} \rightarrow _{6}^{12}\textrm{C} + _{0}^{1}\textrm{n}$
Thus atomic mass is conserved in the reaction.

When the number of nucleons in the nucleus increased then the binding energy per nucleon

0%
decreases continuously with A
0%
increases continuously with A
0%
remains constant with A
0%
first increases with A and then decreases

$_{1}^{2}\textrm{H}$ +

$_{4}^{9}\textrm{Be}$

$\rightarrow$

$X$ +

$_{2}^{4}\textrm{He}$ identify

$X$

0%
$_{3}^{7}\textrm{Li}$
0%
$_{3}^{6}\textrm{Li}$
0%
$_{4}^{7}\textrm{Be}$
0%
$2$ $_{2}^{3}\textrm{He}$

Explanation

The given eq. is:

$_{1}^{2}H+_{4}^{9}Be\longrightarrow X+_{2}^{4}He$

Let X is an atom with atomic no. x and atomic weight y i.e.

$_{x}^{y}X$

For a balanced equation:

$1+4=x+2$

$x=3$

and

$2+9=y+4$

$y=7$

which is best suited to

$_{3}^{7}Li$ out of the given options .

Which of the following reactions is impossible?

0%
$_{2}\textrm{He}^{4}$ + $_{4}\textrm{Be}^{9}$ = $_{0}\textrm{n}^{1}$ + $_{6}\textrm{C}^{12}$
0%
$_{2}\textrm{He}^{4}$ + $_{7}\textrm{N}^{14}$ = $_{1}\textrm{H}^{1}$ + $_{8}\textrm{O}^{17}$
0%
4( $_{1}\textrm{H}^{1}$ ) = $_{2}\textrm{He}^{4}$ + 2( $_{1}\textrm{e}^{0}$ )
0%
$_{3}\textrm{Li}^{4}$ + $_{1}\textrm{H}^{1}$ = $_{4}\textrm{Be}^{8}$

Explanation

In the nuclear reactions atomic mass of nucleons should remained conserved on either sides of reaction or mass of product nuclei should be less than the reacting ones.
A] In first reaction the total atomic mass of reactant nuclei i.e.

$He$ and

$Be\ (9 + 4 = 13)$ is equal to the mass of product nuclei i.e. C and neutron

$(12 + 1 = 13).$
B] Also, in second reaction the total atomic mass of reactant nuclei i.e.

$He$ and

$N (4 + 14 = 18)$ is equal to the mass of product nuclei i.e. proton and oxygen

$(1 + 17 = 18)$ .
C] Similarly in third reaction mass is conserved between

$4$ protons and alpha particle.
D] In this reaction, mass of lithium and proton is less than the isotope of

$Be$ this is not possible since, the light nuclei combines to form the heavier one of equal mass or with some mass defect which is converted into energy. But in given reaction mass of product nucleus is large which is in contradiction.

Two deutrons fuse to form a helium nucleus and energy is released, because the mass of helium nucleus is

0%
equal to that of two deutrons
0%
less than that of two deutrons
0%
more than that of two deutrons
0%
all the above

Explanation

Mass of deuteron nucleus is 2.0141 amu and masss of helium nucleus is 4.0026 amu. When two deuterons fuse to form a helium nucleus some of the mass will be lost and the mass defect is

$\Delta m = (2 \times 2.0141) - 4.0026$

$\Delta m = 4.0282 - 4.0026$

$\Delta m = 0.0256 amu$
This mass is converted into energy and is released.
Hence, two deuterons fuse to form a helium nucleus and energy is released, because the mass of helium nucleus is less than that of two deuterons

The curve between binding energy per nucleon

$(E)$ and mass number

$(A)$ is:

Explanation

We know that the binding energy per nucleon (E) increases upto

$A=56$ then it decreases as A increases. So

$^{56}Fe$ is the most stable element. Thus, option C will represent correct graph.

The mass defect for helium nucleus is

$0.0304$

$a.m.u$ . The binding energy per nucleon of helium nucleus is ________

0%
$28.3$ $MeV$
0%
$7.075$ $MeV$
0%
$9.31$ $MeV$
0%
$200$ $MeV$

Explanation

$1\ amu = 1.67 \times 10^{-27} kg$
Using this and Einsteins relation, we get the energy equivalent to the rest mass as:

$E = mc^{2} = (1.67 \times 10^{-27} \times 0.0304) \times (3 \times 10^{8})^{2} =4.569 \times 10^{-12}J$

Converting this energy in electron volts.

$E = \dfrac{4.569 \times 10^{-12}}{1.6 \times 10^{-19}} = 28MeV$

This is per nucleon of Helium. One Helium nucleus has four nucleons.

Hence, binding energy per nucleon is

$7\ MeV$ .

The nucleus obtained after

$\alpha$ -emission from the nucleus

$_{y}\textrm{A}^{x}$ is

0%
$_{y-2}\textrm{B}^{x-2}$
0%
$_{y+2}\textrm{B}^{x+4}$
0%
$_{y}\textrm{B}^{x}$
0%
$_{y-2}\textrm{B}^{x-4}$

Explanation

The emission of an alpha particle from a nucleus reduces the mass number of the nucleus by

$4$ and the atomic number by

$2$ .

The energy of thermal neutrons is nearly

0%
$0.25$ $eV$
0%
$0.025$ $eV$
0%
$200$ $MeV$
0%
$0.025$ $Joule$

Explanation

A thermal neutron is a free neutron with a kinetic energy of about

$0.025\ eV$ (about

$4.0\times10^{-21}J$ or

$2.4\ MJ/kg$ , hence a speed of

$2.2\ km/s$ ), which is the energy corresponding to the most probable velocity at a temperature of

$290\ K$ .

The particle

$X$ in the following nuclear reaction is

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

$_{1}^{1}\textrm{H}$

$\longrightarrow$

$_{2}^{4}\textrm{He}$ +

$X$

0%
$a$
0%
$n$
0%
$e$
0%
$p$

Explanation

Mass number of

$Li$ is

$7$ .

$X$ should have a mass number of

$4$ and atomic number of

$2$ .

$X$ is a

$\alpha$ particle.

The fusion process is possible at high temperature because at high temperatures

0%
the nucleus disintegrates
0%
molecules disintegrate
0%
atoms become ionised
0%
the nuclei get sufficient energy so as to overcome the Coulomb repulsive force

The particle

$X$ in the following nuclear reaction is

$_{5}^{B}\textrm{10}$ +

$_{2}^{He}\textrm{4}$

$\longrightarrow$

$_{7}^{N}\textrm{13}$ +

$X$

0%
$P$
0%
$a$
0%
$e$
0%
$n$

Explanation

Since the atomic number remains same (i.e

$5+2=7$ ), but mass number is changed, it is a neutron.

The temperature necessary for fusion reaction is

0%
$3\times10^{3}$
0%
$3\times10^{6}$
0%
$3\times10^{2}$
0%
$3\times10^{4}$

The fusion of light elements take place at about the temperatures of about

0%
$30\ ^{0}C$
0%
$100\ ^{0}C$
0%
$10,000\ ^{0}C$
0%
$2\times10\ ^{0}C$

Which of the following is correct?

0%
There are $78$ neutrons in $_{78}{Pt}^{192}$
0%
$_{84}{Po}^{214}$ $\rightarrow$ $_{82}{Pb}^{210}$ + $\beta$
0%
$_{92}{U}^{238}$ $\rightarrow$ $_{90}{Th}^{234}$ + $_{2}{He}^{4}$
0%
$_{90}{Th}^{234}$ $\rightarrow$ $_{91}{Pa}^{234}$ + $_{2}{He}^{4}$

Explanation

After release of helium, there will be decrease in atomic number by

$2$ and mass number by

$4$ .

Let

$E_1$ and

$E_2$ be the binding energies of two nuclei A and B. It is observed that two nuclei of A combine together to form a B nucleus. This observation is correct only if

0%
$E_1 > E_2$
0%
$E_2 > E_1$
0%
$E_2 < 2E_1$
0%
nothing can be said

Explanation

$2A \rightarrow B$
Possible if B is more stable than A

$\implies$ Energy of B is less than two atoms of A

$E_2 < 2E_1$

According to Bohr's Theory of hydrogen atom, the product of the binding energy of the electron in the

$nth$ orbit and its radius in the

$nth$ orbit

0%
Is proportional to $n^2$
0%
Is inversely proportional to $n^3$
0%
Has a constant value of $10.2\space eV-\mathring{A}$
0%
Has a constant value of $7.2\space eV-\mathring{A}$

Explanation

From the formulas attached.
The product

$R(n) \times E(n) = 0.529\cfrac{n^2}{Z} \times 13.6 \cfrac{Z^2}{n^2} eV.-A$

$R(n) \times E(n) = 0.529 \times 13.6 Z eV.-A$

$R(n) \times E(n) = 7.2 eV-A$

$1$ amu is equal to

$1.66\times 10^{-24}$ g.

0%
True
0%
False

Explanation

The atomic mass unit (amu) is equal to

$\frac{1}{12}$ of the mass of one atom of the carbon-12 isotope.

1 mole carbon-12 atoms weigh 12 g.

$6.022 \times 10^{-23}$ carbon -12 atoms weigh

$12g$ .

1 carbon-12 atom weighs

$\dfrac {12}{6.022 \times 10^{-23}} \ g$

$\therefore 1\ amu = \dfrac{1}{12} \times$ mass of one C-12 atom

$= \dfrac {1}{12}$

$\times \dfrac {12}{6.022 \times 10^{-23}}$

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

$1.66 \times 10^{-27} kg$ .

The Sun's mean density is

0%
$1.4\times kg/m^{3}$
0%
$1.4\times10^{3} kg/m^{3}$
0%
$1.4\times10^{3} kg/cc$
0%
$1.4\times kg/cc$

Explanation

Density of water is

$=1000\ kg/m^3$

Density of sun

$=1.4\times 1000=1.4\times 10^3 ms^{-3}$

An element A decays into an element C by a two-step process:

$A\rightarrow B+He_2^4$ and

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

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

Explanation

Conservation of mass no. and charge holds true

$A \rightarrow B$

$Z_B = Z_A -2$

$M_B = M_A -4$

$B \rightarrow C + 2e$

$Z_C = Z_B +2 = Z_A$

$M_C = M_B = M_A-4$
Different mass, but same atomic No.
Therefore, A and C are isotopes.

$_{92}U^{238}$ absorbs a neutron. The product emits an electron. This product further emits an electron. The result is

0%
$_{94}Pu^{239}$
0%
$_{90}Pu^{239}$
0%
$_{93}Pu^{237}$
0%
$_{94}Pu^{237}$

Explanation

$_{92}V^{238}+ n \rightarrow _{92}A^{239}$

$_{92}A^{239} \rightarrow _{93}B^{239} +e$

$_{92}B^{239} \rightarrow _{94}C^{239} +e$
Finding the element C from periodic table

$_{94} Pu^{239}$

The minimum frequency of a

$\gamma$ -ray that causes a deuteron to disintegrate into a proton and a neutron is

$(m_d=2.0141 amu, m_p=1.0078 amu, m_n=1.0087 amu.)$

0%
$2.7\times 10^{20}Hz$
0%
$5.4\times 10^{20}Hz$
0%
$10.8\times 10^{20}Hz$
0%
$21.6\times 10^{20}Hz$

Explanation

$D \rightarrow n +p$

$\Delta m = m_p +m_n - m_d$

$=1.0087 +1.0078 -2.0141 )amu \\ = 2.4 \times 10^{-3} amu$

Energy required

$=\Delta mc^2$

$E/amu = 931\ Mev$

Energy

$= 931(2.4 \times 10^{-3})= 2.2344\ MeV$

$h v = 2.23\ MeV$

$v = 5.4 \times 10^{20} Hz$

An element X decays, first by positron emission and then two

$\alpha$ -particles are emitted in successive radioactive decay. If the product nucleus has a mass number

$229$ and atomic number

$89$ , the mass number of atomic number of element X are

0%
$237, 93$
0%
$237, 94$
0%
$221, 84$
0%
$237, 92$

Explanation

$X \rightarrow e^+ + 2 \alpha + Y$

$\alpha = He^{2+}$
Mass of

$\alpha =4$

Mass no. of

$Y = 229$
By conservation of mass:
Mass no. of

$X = My + 8 = 237$

Conservation of charge:

$\alpha = 2n + 2p$
No. of protons in

$Y = 89$
No. of protons in

$X= 89 + 4+1$ (positron)

$= 94$
Hence, atomic no. of

$X = 94$

During a nuclear fusion reaction

0%
a heavy nucleus breaks into two fragments by itself
0%
a light nucleus bombarded by thermal neutrons breaks up
0%
a heavy nucleus bombarded by thermal neutrons breaks up
0%
two light nuclei combine to give a heavier nucleus and possibly other products

If the Q value of an endothermic reaction is

$11.32 MeV$ , then the minimum energy of the reactant nuclei to carry out the reaction is (in laboratory frame of reference)

0%
$11.32 MeV$
0%
less than $11.32 MeV$
0%
grater than $11.32 MeV$
0%
Data is insufficient

Explanation

From conservation of energy:

Change in energy

$=$ Energy of reactants

$-$ Energy of products

$-$

$Q>0$ (Endothermic)

Therefore, minimum energy of reactants

$>Q=11.32\ MeV$

$^{22}Ne$ nucleus, after absorbing energy, decays into two

$\alpha$ -particles and an unknown nucleus. The unknown nucleus is

0%
nitrogen
0%
carbon
0%
boron
0%
oxygen

Explanation

$^{22}Ne$ decays

$\alpha$ particle =

$He^{2+}$
Mass No = 4

$p=2, n=2$
So, New mass no.

$= 22 - 8 = 14$
Atomic No.

$= 10 -4 = 6$
So, the new element is

$_6C^{14}$

Binding energy per nucleon for

$C^{12}$ is 7.68 MeV and for

$C^{13}$ is

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

$C^{13}$ is

0%
$5.49 MeV$
0%
$8.46 MeV$
0%
$9.45 MeV$
0%
$15.49 MeV$

Explanation

$C^{13} +energy \rightarrow C^{12} +n$
Energy required to remove one neutron = Difference in total binding energy

$= 13 \times 7.74 - 12 \times 7.68$ MeV

$= 8.46 MeV$

Atomic number (Z) of a neutral atom and mass number (A) of an atom are equal to:

(Here n = number of neutrons and p = number of protons):

0%
Z = n and A = n+p
0%
Z = e and A = n+e
0%
Z = p and A = n+p
0%
Z = n and A = p+e

Which of the following is the radio isotope in this pair?

$\underset{6}{12} C, \underset{6}{14} C$

0%
$\underset{6}{12} C$
0%
$\underset{6}{14} C$
0%
Both of them
0%
None of them

Explanation

The

${}_6^{14}C$ has more number of neutrons than protons, so it is the radio isotope in this pair.

Which of the following is the radio isotope in this pair ?

$^{12}_{6}\,C, \, ^{14}_{6}\,C$

0%
$^{14}_{6}\,C$
0%
$^{12}_{6}\,C$
0%
Both of them
0%
None of them

Mass defect of

$1 g$ gives energy equal to

0%
$9 \times 10^{13} \, J$
0%
$5.625\times 10^{32} \, eV$
0%
$2.15\times 10^{10} \, kcal$
0%
$9 \times 10^{6} \, ergs$

Explanation

Given: mass defect,

$\Delta m = 1 g= 10^{-3} kg$

Using Einstein equation,

$E= \Delta m c^2$

$E=10^{-3} \times (3\times 10^8)^2 J = 9\times 10^{13} J$

$E= \dfrac{9\times 10^{13}}{4.18} cal = 2.15 \times 10^{13} cal$

$E= \dfrac{9\times 10^{13}}{1.6\times 10^{-19}} eV = 5.625 \times 10^{32} eV$

$\alpha$ -particle captures an electron. What does it change to?

0%
$He^{2+}$
0%
$He$
0%
$He^{+}$
0%
$\alpha$

Explanation

$\alpha$ particle has two protons and two neutrons and zero electrons. It is written as

${}_2^4He^{2+}$ .

So if it captures an electron, the reaction is :

${}_2^4He^{2+}+e^{-}\rightarrow{}_2^4He^+$

In the nuclear reaction

$H^2+H^2\rightarrow He^3 +X,$

$X$ stands for:

0%
$_{+1}{e^{0}}$
0%
$_0{n^{1}}$
0%
$_{-1}{e^{0}}$
0%
$H^1$

Explanation

The given equation or reaction represents the nuclear fusion reaction

$_1H^2 +_1H^2\rightarrow He^3+ _on^1+hv$

Hence,

$X$ stands for

$_on^1$ .

Which one of them is a radioisotope?

0%
Cobalt-60
0%
Carbon-12
0%
Carbon-14
0%
None of the above

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

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

The mass of a

$^7_3Li$ nucleus is

$0.042 u$ less than the sum of the masses of all its nucleons. What is the binding energy per nucleon of

$^7_3Li$ nucleus ?

0%
$23 MeV$
0%
$46 MeV$
0%
$5.6 MeV$
0%
$3.9 MeV$

Explanation

As,

$1u\equiv 931\ MeV$

So, Binding energy of

${}_3^7Li$ is

$0.042\times931=39.102\ MeV$

Binding energy per nucleon is

$39.102/7=5.59\approx5.6\ MeV$

The mass defect per nucleon is called

0%
binding energy
0%
packing fraction
0%
ionisation energy
0%
excitation energy

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

$X$ . Find the value of

$X/25$ . (Assume that daughter nucleus is in ground state)

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

The

$\alpha-$ decay nuclear reaction:

$^X_Z B \rightarrow ^{X-4}_{Z-2} C + \alpha + Q$

where,

$B$ and

$C$ are the mother and daughter nuclei respectively.

Given:

$Q = 50 MeV, \ E_{alpha} = 48 MeV$

Using:

$Q = E_{\alpha} \bigg(\dfrac{M_{\alpha} + M_{C}}{M_C} \bigg)$

we get,

$Q = E_{\alpha} \bigg(\dfrac{4 + X-4}{X-4} \bigg) = E_{\alpha} \bigg(\dfrac{X}{X-4}\bigg)$

$\therefore 50 = 48 \bigg(\dfrac{X}{X-4} \bigg) \implies X = 100$

Hence,

$\dfrac{X}{25} = \dfrac{100}{25}=4$

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 molecular nucleus is

$X$ . Find value of

${ X }/{ 25 }$ .
(Assume that daughter nucleus is in ground state.)

0%
$2$
0%
$4$
0%
$6$
0%
$8$

Explanation

We have:

${ K }_{ a }=\displaystyle \frac { { m }_{ y } }{ { m }_{ y }+{ m }_{ \alpha } } Q$

$\Rightarrow { K }_{ \alpha }=\displaystyle \frac { A-4 }{ A } Q$

$\Rightarrow 48=\dfrac { A-4 }{ A } \times 50$

$\Rightarrow A=100$

The masses of neutron and proton are

$1.0087 a.m.u.$ and

$1.0073 a.m.u.$ respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass

$4.0015 a.m.u$ , then the binding energy fo the helium nucleus will be: (

$1\ a.m.u. = 931 MeV$ )

0%
$28.4\ MeV$
0%
$20.8\ MeV$
0%
$27.3\ MeV$
0%
$14.2\ MeV$

Explanation

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

$=\Delta \times 931\ MeV\\ =\left[ 2\left( 1.0087+1.0073 \right) -4.0015 \right] \times 931\\ =28.4\ MeV$

If the total binding energies of

$_{ 1 }^{ 2 }{ H, }\ _{ 2 }^{ 4 }{ He, }\ _{ 26 }^{ 56 }{ Fe \ \ \& }\ _{ 92 }^{ 235 }{ U }$ nuclei are

$2.22, 28.3, 492$ and

$1786$ MeV respectively, identify the most stable nucleus of the following.

0%
$_{ 26 }^{ 56 }{ Fe }$
0%
$_{ 1 }^{ 2 }{ H }$
0%
$_{ 92 }^{ 235 }{ U }$
0%
$_{ 2 }^{ 4}{ He }$

Explanation

The binding energy is the energy that is required to break the nucleons apart. The curve of Binding energy per nucleon Vs the Mass number shows the highest binding energy is that of Iron no other is tightly bound. Except the light nuclei having

$BE. 56MeV.$

Hence the most stable nucleon is

$2.22 \ MeV$

If the total binding energies of

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

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

$_{ 26 }^{ 56 }{ Fe }$ &

$_{ 92 }^{ 235 }{ U }$ are

$2.22, 28.3, 492$ and

$1786 MeV$ respectively, identify the most stable nucleus of the following

0%
$_{ 26 }^{ 56 }{ Fe }$
0%
$_{ 1 }^{ 2 }H$
0%
$_{ 92 }^{ 235 }{ U }$
0%
$_{ 4 }^{ 2 }He$

Explanation

$B.{ E }_{ H }=\dfrac { 2.22 }{ 2 } =1.11$

$B.{ E }_{ He }=\dfrac { 28.3 }{ 2 } =7.08$

$B.{ E }_{ Fe }=\dfrac { 492 }{ 56 } =8.78=$ maximum

$B.{ E }_{ U}=\dfrac { 1786 }{ 235 } =7.6$

$_{ 26 }^{ 56 }{ Fe }$ is most stable as it has maximum binding energy per nucleon.

In the process of fission, the binding energy per nucleon

0%
increases
0%
decreases
0%
remains unchanged
0%
increases for mass number $A<56$ nuclei but decreases for mass number $A>56$ nuclei

Explanation

For fission, energy to be realeased

$E = (BE)_{products} - (BE)_{reactants}$
If products have to be more stable than the reactant, the BE per nucleon has to be higher for products.
Hence, it releases the energy and reaction continues.

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 5 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 5 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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