Tag: atomic nuclei

Questions Related to atomic nuclei

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

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $3.2\times { 10 }^{ -13 } J$

  2. $1.6\times { 10 }^{ -13 } J$

  3. $4.8\times { 10 }^{ -13 } J$

  4. $6.4\times { 10 }^{ -13 } J$

Show answer
Correct Option: B
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.$

The rest energy of an electron is

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $510 KeV$

  2. $931 KeV$

  3. $510 MeV$

  4. $931 MeV$

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Correct Option: A
Explanation:

Rest energy of an electron $=m _{ e }{ c }^{ 2 }$
Here $m _{ e }=9.1\times { 10 }^{ -31 }kg$ and $C$ = velocity of light

$\therefore$ Rest energy $=9.1\times { 10 }^{ -31 }\times\left( 3\times { 10 }^{ 8 } \right) ^{ 2 }joule$

                         $=\displaystyle \frac { 9.1\times { 10 }^{ -31 }\times \left( 3\times { 10 }^{ 8 } \right) ^{ 2 } }{ 1.6\times { 10 }^{ -19 } } eV\simeq 510KeV$

1mg of matter convert into energy will give

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $90$ joule

  2. $9\times { 10 }^{ 3 }$ joule

  3. $9\times { 10 }^{ 5 }$ joule

  4. $9\times { 10 }^{ 10 }$ joule

Show answer
Correct Option: D
Explanation:

$E=m{ c }^{ 2 }=10^{ -3 }\times { 10 }^{ -3 }\times { \left( 3\times { 10 }^{ 8 } \right)  }^{ 2 }= 9\times { 10 }^{ 10 }J$

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) $

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $1.5\times { 10 }^{ 6 }$

  2. $2.5\times { 10 }^{ 6 }$

  3. $3\times { 10 }^{ 6 }$

  4. $7.5\times { 10 }^{ 6 }$

Show answer
Correct Option: D
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 $

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 $\displaystyle { C }^{ 13 }$ ?

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. $5.34MeV$

  2. $5.5MeV$

  3. $9.5 MeV$

  4. $9.34MeV$

Show answer
Correct Option: A
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 $97.5-92.16=5.34\ MeV$

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.

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. 30

  2. 22

  3. 35

  4. 26

Show answer
Correct Option: D
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, $\dfrac{0.04\times10^{-6}}{2}=(\dfrac{1}{2})^{n}$ 

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

             $log 2-8log10=-nlog2$ 

             $0.3010-8=-0.3010n$ 

             $n=0.7699/0.3010=25.58$

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 $1.0078  u$.

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. 19.285  MeV

  2. 14.232 MeV

  3. 17.326 MeV

  4. 23.564 MeV

Show answer
Correct Option: C
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 $= 0.0186  u \times 931.5  MeV = 17.326  MeV$

The binding energy of $ _{3}{Li}^{7}$ and $ _{2}{He}^{4}$ are $39.2  MeV$ and $28.24  MeV$ respectively. Which of the following statements is correct?

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. Helium is more stable than lithium.

  2. Lithium is more stable than helium.

  3. Both are equally stable

  4. None of the above

Show answer
Correct Option: A
Explanation:

The nucleons present in $ _{3}{Li}^{7}$ is $7$.
The binding energy per nucleon for lithium is ${39.2}/{7} = 5.6  MeV$
The binding per nucleon for helium is ${28.24}/{4} = 7.06  MeV$
The binding energy per nucleon is the measure of stability of the nuclei. Therefore, helium is more stable than lithium.

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?

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. ${He} _{2}^{4}$

  2. ${U} _{92}^{235}$

  3. $ _{1}{H}^{2}$

  4. $ _{26}{Fe}^{56}$

Show answer
Correct Option: D
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.

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

nuclear reactions nuclear structure nuclei atomic nuclei physics

  1. momentum

  2. mass

  3. energy

  4. all of these

Show answer
Correct Option: A
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