Physics

Nuclear and Atomic Physics

571 Questions

Nuclear and atomic physics explores the components and properties of the nucleus, including isotopes, radioactive decay, and fundamental forces. These concepts are essential for various competitive exams requiring a strong foundation in physics. The provided questions cover structural properties, mass, energy equivalence, and particle interactions.

Nucleus propertiesIsotopes and mass numberAlpha particle scatteringNuclear forcesMass energy equivalenceBeta particle emission

Nuclear and Atomic Physics Questions

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

If 1mg of ${ U }^{ 235 }$ is completely annihilated, the energy liberated is

  1. $\quad 9\times { 10 }^{ 10 } J$
  2. $\quad 9\times { 10 }^{ 19} J$
  3. $\quad 9\times { 10 }^{ 18} J$
  4. $\quad 9\times { 10 }^{ 17} J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$E={ mc }^{ 2 }={ 10 }^{ -6 }\times \left( 3\times { 10 }^{ 8 } \right) ^{ 2 }={ 10 }^{ -6 }\times 9\times { 10 }^{ 16 }=9\times { 10 }^{ 10 }J$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

If an electron and positron annihilate, then the energy released is

  1. $3.2\times { 10 }^{ -13 } J$
  2. $1.6\times { 10 }^{ -13 } J$
  3. $4.8\times { 10 }^{ -13 } J$
  4. $6.4\times { 10 }^{ -13 } J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy of electron $={ m } _{ e }{ c }^{ 2 }$
Energy of positron $={ m } _{ p }{ c }^{ 2 }$
${ m } _{ e }$ = ${ m } _{ p }$, $ c=$ speed of light.

Thus according to conservation of energy released $=2{ m } _{ e }{ c }^{ 2 }$$=2\times 9.1\times { 10 }^{ -31 }\left( 3\times { 10 }^{ 8 } \right) ^{ 2 }=1.6\times { 10 }^{ -13 } Joules.$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

The mass defect in a particular nuclear reaction in 0.3 grams.The amount of energy liberated in kilowatt hour is $\left( Velocity\ of \  light=3\times { 10 }^{ 8 }m/s \right) $

  1. $1.5\times { 10 }^{ 6 }$
  2. $2.5\times { 10 }^{ 6 }$
  3. $3\times { 10 }^{ 6 }$
  4. $7.5\times { 10 }^{ 6 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Mass defect in a nuclear reaction          $\Delta M = 0.3    g  =  3 \times 10^{-4}    kg$

Thus amount of energy released        $E = \Delta M    c^2  =  (3 \times 10^{-4}) \times (3 \times 10^8)^2           J$
$\implies         E =  27  \times 10^{12}      J                                  (1   kWh = 3.6  \times 10^6    J)$
$\therefore         E  = 7.5   \times 10^{6}     kWh $

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

The binding energy per nucleon for $\displaystyle { C }^{ 12 }$ is $7.68 MeV$ and that for $\displaystyle { C }^{ 13 }$ is $7.5 MeV$. How much energy is  required to remove a neutron from $\displaystyle { C }^{ 13 }$ ?

  1. $5.34MeV$
  2. $5.5MeV$
  3. $9.5 MeV$
  4. $9.34MeV$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Total B.E. for $C^{13}$ is $13\times7.5=97.5\ MeV$

Total B.E. for $C^{12}$ is $12\times7.68=92.16\ MeV$
The energy required to remove a neutron from $C^{13}$ is $97.5-92.16=5.34\ MeV$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

When a neutron collides with a quasi free proton, it loses half of its energy on the average in the every collission. How many collisions, on the average, are required to reduce a 2 MeV neutron to a thermal energy df 0.04 eV.

  1. 30

  2. 22

  3. 35

  4. 26
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $E _{0}$ be the initial energy of neutron, the energy of neutron after 1 collision reduces to $E _{0}/2=E _{1}(let)$ i.e. $E _{1}/E _{0}=1/2$. 


After second collision, $E _{2}/E _{0}=(1/2)^{2}$, therefore after $n$ collision.
           $\dfrac{E _{n}}{E _{0}}=(\dfrac{1}{2})^{n}$ 


Here, given $E _{0}=2MeV , E _{n}=0.04eV=0.04\times10^{-6}MeV$ 

Hence, $\dfrac{0.04\times10^{-6}}{2}=(\dfrac{1}{2})^{n}$ 

             $2\times10^{-8}=(\dfrac{1}{2})^{n}$ 

             $log 2-8log10=-nlog2$ 

             $0.3010-8=-0.3010n$ 

             $n=0.7699/0.3010=25.58$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

Find the energy released during the following nuclear reaction.


$ _{1}{H}^{1}  +   _{3}{Li}^{7}  \longrightarrow   _{2}{He}^{4}  +   _{2}{He}^{4}$

The mass of $ _{3}{Li}^{7}$ is $7.0160  u$,  $ _{2}{He}^{4}$ is $4.0026  u$ and proton is $1.0078  u$.

  1. 19.285 MeV

  2. 14.232 MeV

  3. 17.326 MeV

  4. 23.564 MeV
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The mass of the reactant nuclei $= 7.0160 + 1.0078 = 8.0238  u$
The mass of the product nuclei $= 4.0026 + 4.0026 = 8.0052  u$
Mass defect $= \Delta m = 8.0238 - 8.0052 = 0.0186  u$
Energy released $= 0.0186  u \times 931.5  MeV = 17.326  MeV$

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

Katen was studying nuclear physics. There, he collected values of binding energies of $ _{1}{H}^{2},   _{2}{He}^{4},   _{26}{Fe}^{56}$ and $ _{92}{U}^{235}$ and they are $2.22  MeV,  28.3  MeV,  492  MeV$ and $1786  MeV$ respectively. Then, he got a doubt that stability of the nucleus depends on its binding energy, which among the above four is the most stable nucleus?

  1. ${He} _{2}^{4}$
  2. ${U} _{92}^{235}$
  3. $ _{1}{H}^{2}$
  4. $ _{26}{Fe}^{56}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Stability of nucleus $\alpha$ $\cfrac{Binding\;Energy}{Atomic\;mass}$
So, ${ _{ 1 }{ H }^{ 2 } }\rightarrow \cfrac { 2.22 }{ 2 } =1.11,\quad { _{ 2 }{ He }^{ 4 } }\rightarrow \cfrac { 28.3 }{ 4 } =7.075\\ { _{ 26 }{ Fe }^{ 56 } }\rightarrow \cfrac { 492 }{ 56 } =8.7,\quad { _{ 92 }{ U }^{ 235 } }\rightarrow \cfrac { 1786 }{ 235 } =7.6$
So, ${ _{ 26 }{ Fe }^{ 56 } }$ is stable among all four.
Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

In the nuclear reaction, there is a conservation of ______.

  1. momentum

  2. mass

  3. energy

  4. all of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In a nuclear reaction, there may be conversion of some mass into energy. So,both mass and energy are not conserved. It is the momentum which is conserved.a

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

The binding energy per nucleon of $^{16}O$ is $7.97MeV$ and that of $^{17}O$ is $7.75MeV$. The energy in MeV required to remove a neutron from $^{17}O$ is:

  1. $3.52$
  2. $3.64$
  3. $4.23$
  4. $7.86$
  5. $1.68$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

BE per nucleon $^{16}O=7.97MeV$
BE per nucleon $^{17}O=7.75MeV$
$^{17}O\rightarrow { _0n^1}+{^{16}O}$
Energy required to remove neutron
$=17\times 7.75-16\times 7.97$
$=4.23MeV$.

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

The mass defect of a certain nucleus is found to be $0.03$ amu. Its binding energy is:

  1. $27.93$ eV
  2. $27.93$ keV
  3. $27.93$ MeV
  4. $27.93$ GeV
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass defect  $\Delta M = 0.03$ amu
Binding energy  $E _{B} = \Delta Mc^2 = \Delta M\times 931.5 $  MeV
$\therefore \ E _{B} = 0.03\times 931.5$ MeV $ =27.93 $ MeV

Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

Higher the mass defect, higher will be the stability of the nucleus.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Amount of energy required to break the nucleus is known as binding energy of the nucleus.
It depends upon mass defect.
Greater  the mass defect, greater will be the binding energy
Stable nuclei have high binding energies
Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

1 u is equivalent to an energy of

  1. 9.315 MeV

  2. 931.5 KeV

  3. 93.15 MeV

  4. 931.5 MeV
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
According to Einstein mass energy equivalence is represented by
$E=m{ c }^{ 2 }$
Taking $m=1a.m.u$
$=1.66\times { 10 }^{ -27 }㎏$
and $c=3\times { 10 }^{ 8 }㎧$
We get, $E=1.66\times { 10 }^{ -27 }\times { \left( 3\times { 10 }^{ 8 } \right)  }^{ 2 }J$
$=1.49\times { 10 }^{ -10 }J$
As $1MeV=1.6\times { 10 }^{ -13 }J$
$\therefore E=\cfrac { 1.49\times { 10 }^{ -10 } }{ 1.6\times { 10 }^{ -13 } } $
$E=931.25MeV$
Hence,$1a.m.u.=931.25MeV$
Multiple choice nuclear reactions nuclear structure nuclei atomic nuclei physics

Two light nuclei of masses $m _1$ and $m _2 $ are fused to form a more stable nucleus of mass $m _3$ then :-

  1. $m _3 = | m _1 - m _2 | $
  2. $m _3 < ( m _1 + m _2 ) $
  3. $m _3 > ( m _1 - m _2 ) $
  4. $m _3 = | m _1 + m _2 | $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

When two nuclei of masses ${m _1}$ and ${m _2}$ are fused to form a stable nucleus of mass ${m _3}$ and some of the mass is converted in energy.

Therefore,

${m _3} < {m _1} + {m _2}$

Multiple choice chemistry energy production nuclear reactor the nuclear power station isotopes and nuclear chemistry

Consider the following statements A and B and identify the correct statements. 
Statement A : $p-n; p-p; n-n$ forces between nucleons are not equal and charge dependent
statement B : In nuclear reactor the fission reaction will be in accelerating state if the value of neutron reproduction factor is $k>1$

  1. Both A and B are correct

  2. Both A and B are wrong

  3. A is wrong B is correct

  4. A is correct B is wrong
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Nuclear forces is charge independent
i.e. $F _{pp} = F _{pn} = F _{nn}$
If the K>1 the fission reaction will be in accelerating state.
So, correct choice is option C.