Tag: nuclei

Questions Related to nuclei

The rest energy of an electron is

  1. $510 KeV$

  2. $931 KeV$

  3. $510 MeV$

  4. $931 MeV$


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

  1. $90$ joule

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

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

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


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) $

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

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

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

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


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 $\displaystyle { C }^{ 13 }$ ?

  1. $5.34MeV$

  2. $5.5MeV$

  3. $9.5 MeV$

  4. $9.34MeV$


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.

  1. 30

  2. 22

  3. 35

  4. 26


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 $1.0078  u$.

  1. 19.285  MeV

  2. 14.232 MeV

  3. 17.326 MeV

  4. 23.564 MeV


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?

  1. Helium is more stable than lithium.

  2. Lithium is more stable than helium.

  3. Both are equally stable

  4. None of the above


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?

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

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

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

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


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 ______.

  1. momentum

  2. mass

  3. energy

  4. all of these


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

The difference between a nuclear reactor and an atomic bomb is that

  1. no chain reaction takes place in nuclear reactor while in the atomic bomb there is a chain reaction

  2. the chain reaction in nuclear reactor is controlled

  3. the chain reaction in nuclear reactor is not controlled

  4. no-chain reaction takes place in atomic bomb while it takes place in nuclear reactor


Correct Option: B
Explanation:

The chain reaction in nuclear reactor is controlled 

Both in nuclear reactor and atomic bomb nuclear fission takes place. But in nuclear reactor controlled fission chain reaction takes place while in atomic bomb chain reaction is uncontrolled.