Question

It is found that if a neutron suffers an elastic collinear collision with deuterium at rest, fractional loss of its energy is $p_d$; while for its similar collision with carbon nucleus at rest, fractional loss of energy is $p_C$. The values of $p_d$ and $p_C$ are respectively:

• A. $\left(0,0\right)$
• B. $\left(0,1\right)$
• C. $\left(0.89,0.28\right)$
• D. $\left(0.28,0.89\right)$

Answer 1

The correct option is C

$\left(0.89,0.28\right)$

Step 1: Given information

1. A neutron suffers an elastic collinear collision with deuterium at rest, and fractional loss of its energy is $p_d$.
2. A similar collision with carbon nucleus at rest, fractional loss of energy is $p_C$.
3. $m,2m$ are masses of two body.
4. $v_{o}istheintialvelocitybeforecollision$
5. $V_1,V_{2}arevelocityaftercollision$

Step 2: Draw the required diagram

Step 3: Calculate the value of $p_d$

L.M.C

$$\begin{gathered} m{V}_1+2m{V}_2=m{V}_0 \\ {V}_1+2V_2=V0\dots \left(1\right) \\ {V}_2–V_1=V_0\dots \left(2\right) \end{gathered}$$

On solving equation (1) and equation (2),

$V_2=\frac{2V_0}{3}and{V}_1=\frac{V_0}{3}$

Final $K.E.$ of neutron $=\left(\frac{1}{2}\right)m{V_1}^2$

$K.E.=\left(\frac{1}{2}\right)m{\left(\frac{V_0}{3}\right)}^2$

$K.E.=\frac{1}{9}\left(\frac{1}{2}m{V_o}^2\right)$

Loss of $K.E.$,

$K.E.=\frac{8}{9}\left(\frac{1}{2}m{V_o}^2\right)$

Step 4: Calculate the value of $p_C$

Fractional loss $p_d=\left(\frac{8}{9}\right)=0.89$

Similarly collision between $NandC$ ,

$m{V}_1+12m.V_2=m{V}_0$

$V_1+12V_2=V_0\dots \left(1\right)$

$V_1-V_2=V_0\dots \left(2\right)$

On Solving equation (1) and equation (2),

$V_2=\frac{2V_0}{13}$

And $V_1=\frac{11V_0}{13}$

Final $K.E.=\frac{1}{2}m\left[\frac{11{V_o}^2}{13}\right]$

$\Rightarrow K.E.=\frac{121}{169}\left[\frac{1}{2}m{V_o}^2\right]$

Loss of $K.E.=\frac{48}{169}\left[\frac{1}{2}m{V_o}^2\right]$

Fractional loss$P_c=\left(\frac{48}{169}\right)=0.28$

Hence, the correct option is (c).