The Correct Answer is Option (C)

Linear momentum is conserved

so, $p_{M^{\prime }/3}=p_{2M^{\prime }/3}$

so, $\frac{{\lambda }_{M^{\prime }/3}}{{\lambda }_{2M^{\prime }/3}}=\frac{1}{1}$

The Correct Answer is Option (C)

$\therefore$$R=R_0{A}^{\frac{1}{3}}$

$\Rightarrow \frac{R_1}{R_2}={\left(\frac{A_1}{A_2}\right)}^{\frac{1}{3}}={\left(\frac{4}{3}\right)}^{\frac{1}{3}}$

$\therefore$ Density ratio, $\frac{{\rho }_1}{{\rho }_2}=\frac{A_1/V_1}{A_2/V_2}$

$=\left(\frac{A_1}{A_2}\right)\times {\left(\frac{R_2}{R_1}\right)}^3$

$=1:1$

The Correct Answer is Option (B)

We know that

$R=R_0{A}^{1/3}$

$\Rightarrow \underset{y}{\underbrace{\ln \left(\frac{R}{R_0}\right)}}=\underset{m}{\underbrace{\frac{1}{3}}}\underset{x}{\underbrace{\ln (A)}}$

$\Rightarrow$ Straight line

The Correct Answer is Option ()

$\lambda =\frac{h}{p}$

both the particles will move with momentum same in magnitude & opposite in direction.

So De-Broglie wavelength of both will be same i.e. ratio 1 : 1

The Correct Answer is Option (C)

$R=(1.3\times {10}^{-15})A^{\frac{1}{3}}$

We know, $m=pV$

$\Rightarrow$ $p=\frac{m}{V}$

$\Rightarrow$ $p=\frac{m_{p}A}{\frac{4}{3}\pi R^3}$

$p=\frac{m_{p}A}{\frac{4}{3}\pi \times {(1.3\times {10}^{-15})}^{3}A}$

$p\approx {10}^{17}kg/m$