MCQ Questions for Class 12 Medical Physics Nuclei Quiz 1

For a nucleus to be stable, the correct relation between neutron number N and proton number Z is

0%
N>Z
0%
N=z
0%
N<Z
0%
$N \geqslant Z$

Let

$u$ denote one atomic mass unit. One atom of an element of mass number

$A$ has mass exactly equal to

$Au$

0%
for any value of $A$
0%
only for $A=1$
0%
only for $A=12$
0%
for any value of $A$ provided the atom is stable

In the nuclear reaction ;

$_{92}U^{238}\rightarrow _{z}Th^{A}+_{2}He^{4}$ the values of A and Z are:

0%
A=230,Z=8
0%
A=234,Z=90
0%
A=228, Z=94
0%
A=232, Z=

If half-life of a radioactive substance is

$60$ minutes, then the percentage decay in

$4$ hours is?

0%
$50\%$
0%
$71\%$
0%
$85\%$
0%
$93.7\%$

In the nuclear reaction :

$X(n, \alpha)_3Li^7$ the term

$X$ will be 3

0%
$_5 B ^{10}$
0%
$_5B^9$
0%
$_5B^{11}$
0%
$_2He^4$

The relation between

$U(r)$ and r for diatomic molecule is given as

$U(r)=\dfrac{a}{r^{12}} - \dfrac{b}{r^5}$
The energy of dissociation of the molecule if given as

0%
$\dfrac{b^2}{4a^2}$
0%
$\dfrac{b^2}{4a}$
0%
$\dfrac{4a^2}{b^2}$
0%
$\dfrac{4a}{b^2}$

The order of magnitude of radius of nucleus is ___________.

0%
$10^{-15} m$
0%
$10^{15} m$
0%
$10^{-10} m$
0%
$10^{10} m$

Explanation

$10^{-15}$ m

The order of magnitude of nuclear radii is 1 fermi = $10^{-15}$ meters.

Fusion reactions take lace at high temperature because

0%
atoms are ionized at high temperature
0%
molecules break up at high temperature
0%
nuclei break up at high temperature
0%
kinetic energy is high enough to overcome repulsion between nuclei

Explanation

Extremely high temps needed for fusion because

$K.E.$ should large enough to overcome repulsion between nuclei.

In the options below which one of the isotope of the uranium can cause fission reaction?

0%
$U_{234}$
0%
$U_{235}$
0%
$U_{237}$
0%
$U_{238}$

$O^{19}\longrightarrow F^{19}+e+\bar v$
In this decay, the rest mass energy of

$O^{19}$ and

$F^{19}$ are

$17692.33\ MeV$ and

$17687.51\ MeV$ respectively. The

$Q$ factor of the decay is :

0%
$4.82\ MeV$
0%
$7\ MeV$
0%
$17.69\ MeV$
0%
$none\ of\ these$

A parent nucleus

$^{m}_{1}p$ decays into a daughter nucleus

$D$ through

$\alpha$ emission in the following way

$^{m}_{1}p\rightarrow D+\alpha$ The subscript and superscript on the daughter nucleus

$D$ will be written as

0%
$^{m}_{n}D$
0%
$^{m+4}_{n}D$
0%
$^{m-4}_{n}D$
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$^{m-4}_{n-2}D$

The appreciable radioactivity of uranium minerals is mainly due to:

0%
An uranium isotope of mass number 235
0%
A thorium isotope of mass number 232
0%
Actinium
0%
Radium

When two deuterium nuclei fuse together to form a tritium nucleus, we get a

0%
neutron
0%
deuteron
0%
alpha particle
0%
proton

Explanation

Here the nuclear reaction :

$2 _1^2H(deutron) \rightarrow _1^3H (tritium)$

For balancing the reaction :

$2 _1^2H \rightarrow _1^3H+ _1^1H(proton)$

Thus, we will get a proton in this reaction.

$M_{0}$ is the mass of an oxygen isotope

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

$M_{N}$ are the masses of a proton and a neutron respectively, the nuclear binding energy of the isotope is

0%
$8{M_p}c^2$
0%
$(-M_{0}+8M_{p}+9M_{N})c^{2}$
0%
$M_{0}c^{2}$
0%
$(M_{0}-17M_{N})c^{2}$

Explanation

$BE = (8M_{p}+9M_{N} -M_{0})C^{2}$

Assume that a neutron breaks into a proton and an electron. The energy released during this process is
(Mass of neutron

$=1.6725\times 10^{-27} kg$ , Mass of proton

$=1.6725\times 10^{-27} kg$ ,Mass of electron

$=9\times 10^{-31} kg$ )

0%
$0.50625$ $\mathrm{MeV}$
0%
$7.10$ $\mathrm{MeV}$
0%
$6.30$ $\mathrm{MeV}$
0%
$5.40$ $\mathrm{MeV}$

Explanation

$n\rightarrow p+ e$

$\Delta m= (m_{p}+m_{e})-m_n$

$\Delta m=-9 \times 10^{-31}kg$
Energy released

$=9 \times 10^{-31} \times (3\times 10^8)^2J$

$=\dfrac{9 \times 10^{-31} \times (3\times 10^8)^2}{1.6\times 10^{-13}}$

$=0.50625\ \mathrm{MeV}$

If the binding energy per nucleon in

$_{7}^{}\textrm{3}Li$ and He nuclei are

$5.60 MeV$ and

$7.06 MeV$ respectively, then in the reaction

$p+_{3}^{}\textrm{7}Li\rightarrow _{2}^{}\textrm{4}$ He, binding energy is

0%
39.2 MeV
0%
28.24 MeV
0%
17.28 MeV
0%
1.46 MeV

Explanation

Since,

$p+_3Li^7 \rightarrow 2_2He^4$

Thus, proton energy = BE of He atom

$-$ BE of Li of atom.

$E_p=(2\times 4 \times 7.06 ) -(7\times 5.60)=17.28 MeV$ where 4 is the no of nucleon of He and 7 is the no of nucleon of Li.

The binding energy per nucleon of deuteron

$(_{1}\mathrm{H}^{2})$ and helium nucleus

$(_{2} \mathrm{He}^{4} )$ is

$1.1$

$\mathrm{M}\mathrm{e}\mathrm{V}$ and

$7$

$\mathrm{M}\mathrm{e}\mathrm{V}$ respectively. If two deuteron nuclei reacts to form a single helium nucleus, then the energy released is

0%
$13.9\mathrm{M}\mathrm{e}\mathrm{V}$
0%
$26.9 \mathrm{M}\mathrm{e}\mathrm{V}$
0%
$23.6 \mathrm{M}\mathrm{e}\mathrm{V}$
0%
$19.2 \mathrm{M}\mathrm{e}\mathrm{V}$

Explanation

Energy required to break

$4$ nucleons in the two deuterons

$=4\times 1.1\ MeV$

Energy released in formation of

$4$ nucleons in the single helium nucleus

$=4\times 7\ MeV$

Hence, total energy released

$=(28-4.4)MeV=23.6\ MeV$

The binding energy per nucleon for the parent nucleus is

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

$\mathrm{E}_{2}$ .Then

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

In the options given below, let

$E$ denote the rest mass energy of a nucleus and

$'n\ '$ a neutron. The correct option is

0%
$\mathrm{E}(_{92}^{236}\mathrm{U})>\mathrm{E}(_{53}^{137}\mathrm{I})+\mathrm{E}(_{39}^{97}\mathrm{Y})+2\mathrm{E}(\mathrm{n})$
0%
$\mathrm{E}(_{92}^{236}\mathrm{U})<\mathrm{E}(_{53}^{137}\mathrm{I})+\mathrm{E}(_{39}^{97}\mathrm{Y})+2\mathrm{E}(\mathrm{n})$
0%
$\mathrm{E}(_{92}^{236}\mathrm{U})<\mathrm{E}(_{56}^{140} \mathrm{B}\mathrm{a} )+\mathrm{E}(_{36}^{94} \mathrm{K}\mathrm{r} )+2\mathrm{E}(\mathrm{n})$
0%
$\mathrm{E}(_{92}^{236}\mathrm{U})=\mathrm{E}(_{56}^{140} \mathrm{B}\mathrm{a} )+\mathrm{E}(_{36}^{94} \mathrm{K}\mathrm{r} )+2\mathrm{E}(\mathrm{n})$

$M$ is atomic weight ,

$A$ is mass number then

$\dfrac{M-A}{A}$ represents :

0%
Mass defect
0%
Packing fraction
0%
Binding energy
0%
Chain reaction

Explanation

Packing fraction is the mass defect per nucleon i.e. elementary particle in the nucleus. The difference between atomic weight and mass number i.e. mass of elementary particle in the nucleus is known as mass defect. Hence,

$p = \dfrac{\Delta m}{A}$

$p = \dfrac{M - A}{A}$

$4_{1}H^{1} \rightarrow _{2}He^{4}+2e^{+} + 26 MeV$
The above reaction represents

0%
Fusion
0%
Fission
0%
$\beta$ -decay
0%
$\gamma$ -decay

Explanation

Overall reaction of fusion is given below

$4 _1^1H \rightarrow _2^4He + 2(_1^0e) + 26MeV$

Reaction of fission is

$_{92}^{235}U + _0^1n \rightarrow _{56}^{141}Ba + _{36}^{92}Kr + 3 _0^1n + Energy$

$\beta$ - decay equation is

$_{Z}^{A}X \xrightarrow [ ]{minus e^-} _{Z+1}^AY$

$\Upsilon$ - decay equation is

$_Z^AX \rightarrow -Z^AX$

A nucleus with mass number

$220$ initially at rest emits an

$\alpha$ -particle. lf the

$\mathrm{Q}$ value of the reaction is

$5.5$

$\mathrm{M}\mathrm{e}\mathrm{V}$ , calculate the kinetic energy of the

$\alpha$ -particle.

0%
$4.4 \mathrm{M}\mathrm{e}\mathrm{V}$
0%
$5.4 \mathrm{M}\mathrm{e}\mathrm{V}$
0%
$5.6 \mathrm{M}\mathrm{e}\mathrm{V}$
0%
$6.5 \mathrm{M}\mathrm{e}\mathrm{V}$

Explanation

Let the velocity of alpha particle be

$v$ .

Mass of the alpha nucleus is 4amu.

From conservation of momentum,

$216(v_{atom})=4v$

$\implies v_{atom}=\dfrac{v}{54}$

Total energy

$=Q=5.5MeV=\dfrac{1}{2}(4)v^2+\dfrac{1}{2}(216)(\dfrac{v}{54})^2$
Solving gives KE of alpha particle

$=\dfrac{1}{2}(4)v^2=5.4MeV$

The

$\beta$ -decay process, discovered around 1900, is basically the decay of a neutron

$(n)$ . In the laboratory, a proton

$(p)$ and an electron

$(e^{-})$ are observed as the decay products of the neutron. Therefore, considering the decay of a neutron as a two-body decay process, it was predicted theoretically that the kinetic energy of the electron should be a constant. But experimentally, it was observed that the electron kinetic energy has a continuous spectrum. Considering a three-body decay process, i.e.

$n\rightarrow p+e^{-}+\overline{v}_{e}$ , around 1930, Pauli explained the observed electron energy spectrum. Assuming the anti-neutrino

$(\overline{v}_{e})$ to be massless and possessing negligible energy, and the neutron to be at rest, momentum and energy conservation principles are applied. From this calculation, the maximum kinetic energy of the electron is

$0.8\times 10^{6}eV$ . The kinetic energy carried by the proton is only the recoil energy.
If the anti-neutrino had a mass of 3

$eV/c^{2}$ (where

$c$ is the speed of light) instead of zero mass, what should be the range of the kinetic energy,

$K$ , of the electron?

0%
$0\leq K\leq 0.8\times 10^{6}eV$
0%
$3.0 eV\leq K\leq 0.8\times 10^{6}cV$
0%
$3.0 eV \leq K<0.8\times 10^{6}eV$
0%
$0\leq K<0.8\times 10^{6}eV$

Explanation

$K$ can't be equal to

$0.8 \times 10^6 eV$ as anti-neutrino must have some energy

Assume that the nuclear binding energy per nucleon (B/A) versus mass number (A) is as shown in the figure. Use this plot to choose the correct choice(s) given below.

0%
Fusion of two nuclei with mass numbers lying in the range of $1 < A < 50$ will release energy
0%
Fusion of two nuclei with mass numbers lying in the range of $51 < A < 100$ will release energy
0%
Fission of a nucleus lying in the mass range of $100 < A < 200$ will release energy when broken into two equal fragments
0%
Fission of a nucleus lying in the mass range of $200 < A < 260$ will release energy when broken into two equal fragments

Explanation

BE per nucleon is more for the products than the reactants.
As we see the plot, for A

$\gt$ BE is less than that for the range

$100 \lt A \lt 200$
Hence, fission will happen for the nuclei in the range

$200$ to

$260.$
Similarly the reactants for fusion will have mass number between 51 and 100 so that the product has mass number between

$100$ and

$200$ .

The nucleus finally formed in fusion of protons in proton-proton cycle is that of :

0%
Heavy hydrogen
0%
Carbon
0%
Helium
0%
Lithium

Explanation

Overall reaction in proton-proton cycle

$4 _1^1H \rightarrow _2^4He + 2(_1^0e) + Q$
(Proton) (Helium nucleus) (Positron) (energy)
So, finally Helium nucleus is formed.

In the core of nuclear fusion reactor, the gas becomes plasma because of

0%
strong nuclear force acting between the deuterons
0%
Coulomb force acting between the deuterons
0%
Coulomb force acting between deuteron-electron pairs
0%
the high temperature maintained inside the reactor core

The source of stellar energy is

0%
Nuclear fission
0%
Nuclear fusion
0%
Nuclear fission & fusion
0%
Nuclear decay

After losing two electrons, an atom of Helium becomes equivalent to

0%
$\alpha$ -particle
0%
$\beta$ -particle
0%
$\gamma$ -particle
0%
Deuterium nucleus

Explanation

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

$_{2}^{4}He^{+2}$ is alpha particle. Because it has charge equal to +2e and mass is four times the mass of one proton.

The Binding energy per nucleon of

$^7_3Li$ and

$_2^4He$ nucleon are

$5.60 MeV$ and

$7.06 MeV$ , respectively. In the nuclear reaction

$_3^7Li+_1^1H\rightarrow _2^4He+^4_2He+Q$ , the value of energy

$Q$ released is

0%
$19.6\ MeV$
0%
$-2.4\ MeV$
0%
$8.4\ MeV$
0%
$17.3\ MeV$

Explanation

$_4Li^7 + _1H^1 \rightarrow _2(_2He^4)$

BE of products

$= (5.6 MeV) \times 7 + 0$

$= 39.2 MeV$

$E = -39.2\ MeV$

BE of reactant

$= (7.06) \times 4 \times 2$

$= 56.48 MeV$

$E_f =- 56.48\ MeV$

As nuclear energy decreases, so some energy will be released

$Q_{release}=E_i-E_f=(-39.2)-(-56.48)=17.28\ MeV$

When a uranium isotope

$^{235}_{92}U$ is bombarded with a neutron, it generates

$^{89}_{36}Kr$ , three neutrons and :

0%
$^{91}_{40}Zr$
0%
$^{101}_{36}Kr$
0%
$^{103}_{36}Kr$
0%
$^{144}_{56}Ba$

Explanation

$_{92}U^{235} + \ _on^1 \to \ _QX^P + _{36}Kr^{89} + 3\ _on^1+ Q(energy)$

$\sum$ Atomic number on

$LHS = 89 + 0 = 89 + 3$

$\sum$ Atomic number on

$RHS = Q + 36 + 3 \times 0$

$\therefore LHS = RHS$

$Q+36 = 92$

$Q = 56$

also,

$\sum$ atomic mass number on

$LHS = 235 + 1 = 236$

$\sum$ Atomic mass number on

$RHS=P + 89 + 3 \times 1$

$\therefore LHS = RHS$

$P + 92= 236$

$P = 144$

So, the element is

$_{56}Ba^{144}$

0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
0%
Assertion is correct but Reason is incorrect
0%
Both Assertion and Reason are incorrect

Explanation

Nuclei having mass number

$A=60$ have the highest binding energy per nucleon.

Thus those are most stable.

Although the Reason is correct, it is not the correct explanation for Assertion.

$\displaystyle { S }^{ 32 }$ absorbs energy and decays into which element after two

$\displaystyle \alpha$ -emissions?

0%
Carbon
0%
Aluminium
0%
Oxygen
0%
Magnesium

Explanation

Two alpha emissions can be written as:

$^{32}_{16}S\rightarrow ^{24}_{12}Mg+2^4_2He$

Thus, sulphur decays into magnesium.

Find

$BE$ per nucleon of

$^{56}Fe$ where

$m(^{56}Fe) = 55.936u\ m_{n} = 1.00727u, m_{p} = 1.007274\ u$ .

0%
$477.45\ MeV$
0%
$8.52\ MeV$
0%
$577\ MeV$
0%
$10.52\ MeV$

Explanation

We know that ,

Binding Energy ,

$BE=\Delta mc^2$ where

$\Delta m$ is the mass defect

Here ,

$\Delta m =\text{Mass of nucleon- Mass of nucleus}= 26\ m_p + 30\ m_n - m^{56}Fe$

Hence

$BE = [26\ m_p + 30\ m_n - m^{56}Fe]c^{2}$

Where ,

$m_{n} = 1.00866\ u$

$m_{p} = 1.00727\ u$

$BE = [26\times 1.00727 + 30 \times 1.00866 - 55.936]\times 931$

$= 0.51282\times 931$

$= 477.435\ MeV$

Binding energy per nucleon

$= \dfrac {477.435}{56}MeV = 8.52\ MeV$ .

Which of the following is wrong statement about binding energy ?

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

The difference between the mass of a nucleus and the combined mass of its nucleons is :

0%
zero
0%
positive
0%
negative
0%
zero, positive or negative

Explanation

We know that mass defect

$=$ combined mass of nucleons

$-$ mass of the nucleus.

Since mass defect is always positive quantity so the difference of nucleus and the combined mass of its nucleons will be negative. The combined mass is greater than the mass of nucleus.

What parameter is used to measure the stability of a nucleus?

0%
Average binding energy
0%
No. of protons
0%
No. of neutrons
0%
No. of electrons

The binding energies of a deutron and an

$\alpha$ -particle are 1.125, 7.2 MeV/nucleon respectively. Which is more stable of the two?

0%
deuteron
0%
$\alpha$ -practicle
0%
both
0%
sometimes deutron and sometimes $\alpha$ -particle

What is the average binding energy per nucleon over a wide range ?

0%
8MeV
0%
8.8MeV
0%
7.6MeV
0%
1.1MeV

Explanation

The average binding energy per nucleon is just the total binding energy divided by the number of nucleons.
If we consider

$Na$ atom, its binding energy is 194 MeV.

Its binding energy per nucleon is given by

$\dfrac {194 MeV} {24} = 8.08 MeV$

When the number of nucleons in a nucleus 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 in mass number

Explanation

The binding energy is the energy released when a nucleus is assembled from its constituent nucleons. It is thus a measure of the amount of energy held within the bonds of the atom and corresponds to the energy required to be put in again to pull the nucleons apart. Hence, the energy equivalent of the mass-defect is called the binding-energy of the nucleus.

The larger the nucleus, the greater is the internal repulsive forces due to the greater number of protons and less energy must be applied to remove a nucleon from the nucleus, hence the binding energy is lower. The greater the binding energy, the more stable the atom is.

This variation in the binding energy per nucleon (

$\cfrac{BE}{A}$ ) is easily seen when the average

$\cfrac{BE}{A}$ is plotted versus atomic mass number (A), as shown in the figure, as the atomic mass number increases i.e. the number of particles in a nucleus increases, the total binding energy also increases first and then decreases for A > 56.

Bombardment of Beryllium by alpha particles resulted in the discovery of :

0%
Proton
0%
Nucleus
0%
Neutron
0%
Positron

Explanation

Bombardment of Beryllium by alpha particles reaction is given as below:

$_4^9Be + _2^4He \rightarrow [_6^{13}C] \rightarrow _6^{12}C + _0^1n$
So, correct choice is option C.

The short range attractive nuclear forces that are responsible for the binding of nucleons in a nucleus are supposed to be caused by the role played by the particles called

0%
Positron
0%
m-Meson
0%
K-Meson
0%
$\pi$ - Meson

Explanation

The nuclear force between a neutron and proton is the result of the exchange of charged mesons

$({\pi}^+{\pi}^-)$ between them.
So, correct answer is

$\pi$ -Meson.

The percentage of mass lost during nuclear fusion is

0%
0.1%
0%
0.4%
0%
0.5%
0%
0.65%

Fusion reactions take place at about :

0%
$3\times 10^{2}$ K
0%
$3\times 10^{3}$ K
0%
$3\times 10^{4}$ K
0%
$3\times 10^{6}$ K

Explanation

Fusion reaction takes place at

$10^7k$

So, the correct choice is

$3\times 10^6k$

In an endothermic reaction the binding energies of reactants and products are

$e_{1}$ ,

$e_{2}$ respectively, then

0%
$e_{1}$ < $e_{2}$
0%
$e_{1}$ = $e_{2}$
0%
$e_{1}$ > $e_{2}$
0%
$e_{1}$ $\geq$ $e_{2}$

If the nuclei of masses

$X$ and

$Y$ are fused together to form a nucleus of mass

$m$ and some energy is released, then

0%
$X+Y=m$
0%
$X+Y< m$
0%
$X+Y> m$
0%
$X-Y=m$

Explanation

Since some energy is released, so one part of total mass of reactants is converted into energy
Therefore, total mass of reactants

$X + Y$
is more than the mass of nucleus (

$m$ )
So, we get

$X + Y > m$

The critical mass of a fissionable material is

0%
0.1 kg
0%
the minimum mass needed for chain reaction
0%
the rest mass equivalent to 1020 joules
0%
0.5 kg

In an energy emitting nuclear reaction the binding energies of reactants and products are

$e_{1}$ ,

$e_{2}$ respectively.Then which is correct of the following ?

0%
$e_{1}$ < $e_{2}$
0%
$e_{1}$ = $e_{2}$
0%
$e_{1}$ > $e_{2}$
0%
$e_{1}$ $\geq$ $e_{2}$

In carbon cycle of fusion, 4 protons combine to yield one alpha particle and

0%
one positron
0%
two positrons
0%
ten positrons
0%
three positrons

Explanation

Carbon cycle [Carbon - Nitrogen Cycle] :
It consists of following reactions :

$_6^{12}C + _1^1H \Rightarrow _7^{13}N + Q_1$

$_7^{13}N \Rightarrow _6^{13}C + (_1^0e) + Q_2$

$_6^{13}C + _1^1H \Rightarrow _7^{14}N + Q_3$

$_7^{14}N + _1^1H \Rightarrow _8^{15}O + Q_4$

$_8^{15}O \Rightarrow _7^{15}N + (_1^0e) + Q_5$

$_7^{15}N + _1^1H \Rightarrow _6^{12}C+ _2^4He + Q_6$
On adding all these equations, we get

$4_1^1H \rightarrow _2^4He + 2(_1^0e) + Q$
(proton) (

$\alpha -$ particle) (positron) (energy)
where,

$Q = Q_1+Q_2+Q_3+Q_4+Q_5+Q_6$ the value of Q as calculated from mass defect is 26.7 MeV.

Why high temperature is required for Nuclear fusion ?

0%
All nuclear reactions absorb heat
0%
The particles can not come together unless they are moving rapidly
0%
The binding energy must be supplied from an external source
0%
The mass defect must be supplied

Explanation

Fusion reaction takes place at temperatures around

$10^7k$ . It requires this high temperature so that nucleus start moving at rapidly speed, which in turn increases their kinetic, so that they overcome the repulsion between them and can come together.

For a 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

Explanation

In fast chain reaction neutron released in previous fission again strikes

$^{235}U$ , So size of

$^{235}U$ block should be greater than it's critical size.

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

CBSE Questions for Class 12 Medical Physics Nuclei Quiz 1 - MCQExams.com

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Practice Class 12 Medical Physics Quiz Questions and Answers

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