Q: A 1.0\ MeV neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator molecule. Number of collision it must be more to reach thermal energy will be \left(Take, \dfrac {3}{2}k_{B}T=0.040\ eV\right)

\begin{array}{l}\text { Let "n" be required number } \\\text { of collision } \\\text { Given } \quad \frac{3}{2} k_{B} T=0.04 \\(0.04) n=1 \\\qquad n=25\end{array}

Question 1

Two protons are at  10  A distance from each other. The nuclear force and electrostatic force between them is  F_n  and  F_e  respectively. Thus:

Accepted Answer

Nuclear forces are the strongest forces in the universe, the nucleus of an atom remains stable because of the presence of nuclear forces between the protons.However, nuclear forces are extremely short range and is experienced by the nucleons only in the femtometer ( 10^{-15} ) range. At the angstrom range nuclear forces are negligible compared to electrostatic force.

Question 2

Why is a  _{ 2 }^{ 4 }{ He }  nucleus more stable than a  _{ 3 }^{ 4 }{ Li }  nucleus?

Accepted Answer

All neutrons attract each other with the same strong nuclear force. So, the strong nuclear force holds together three protons and one neutron  \left( _{ 3 }^{ 4 }{ Li } \right)   just as vigorously as it holds together two protons and two neutrons  \left( _{ 2 }^{ 4 }{ He } \right)  . Specifically, protons repel other protons due to the positive charge over them. This repulsion tries to make a nucleus fly apart. Since  \left( _{ 2 }^{ 4 }{ He } \right)   contains only two protons, the attractive strong nuclear forces overcome the repulsion of the protons. Hence, the nucleus holds together. But in  \left( _{ 3 }^{ 4 }{ Li } \right)  , the mutual repulsion of the three protons overcomes the strong nuclear attractions and the nucleus falls apart (or undergoes radioactive decay into a more stable nucleus). Therefore, the answer will be (c)

Question 3

Question 4

A radioactive material has a half life of  1  minute. If one of the nuclei decays now, when will the next one decay?

Accepted Answer

Radioactive decay is a random process. No matter how long a particular nuclei has existed or what is the half life of material, it is impossible to predict when a particular nuclei will decay.Also that's the reason that we have to calculate mean life of a material because we can't say anything about the exact lifetime of a radioactive material.Hence, the nuclei can decay at anytime.

Question 5

Which sample,  A  or  B  shown in figure has shorter mean-life?

Accepted Answer

Initially at  t=0  from figure given { \left( \cfrac { d{ N }_{ 0 } }{ dt }  ight)  }_{ A }={ \left( \cfrac { d{ N }_{ 0 } }{ dt }  ight)  }_{ B } So  {({N}_{0})}_{A}={({N}_{0})}_{B} ie initially both samples has equal number of radioactive atoms. Considering at any instant  t=t  from figure. { \left( \cfrac { d{ N }_{ 0 } }{ dt }  ight)  }_{ A }>{ \left( \cfrac { d{ N }_{ 0 } }{ dt }  ight)  }_{ B }\quad  { \lambda  }_{ A }{ N }_{ A }={ \lambda  }_{ B }{ N }_{ B } { N }_{ A }>{ N }_{ B } { \lambda  }_{ A }{ 	au  }_{ B } So mean life time of sample  A  is greater than of  B

Question 6

A  1.0\ MeV  neutron emitted in a fusion loses one half of its kinetic energy in each collision with a moderator  molecule. Number of collision it must be more to reach thermal energy will be  \left(Take, \dfrac {3}{2}k_{B}T=0.040\ eV\right)

Accepted Answer

\begin{array}{l}\text { Let &#34;n&#34; be required number } \\text { of collision } \\text { Given } \quad \frac{3}{2} k_{B} T=0.04 \(0.04) n=1 \\qquad n=25\end{array}

Nuclei