In a scattering experiment, a particle of mass 2m collides with another particle of mass m, which is initially at rest. Assuming the collision to be perfectly elastic, the maximum angular deviation θ of the heavier particle, as shown in the figure, in radians is:

Correct Option: The correct answer is $\pi /6$.

Analysis of the Figure:
As shown in the provided schematic diagram, a projectile particle of mass $2m$ moves horizontally to the right with an initial velocity and collides with a target particle of mass $m$ that is initially at rest. After the collision, the heavier particle of mass $2m$ is scattered upwards at an angle $\theta$ relative to its original line of motion, while the lighter particle of mass $m$ moves downwards and to the right.

Step-by-Step Mathematical Derivation:

Let the initial velocity of the heavier particle ($m_1=2m$) be $\overrightarrow{u}=u\hat{i}$, and let the target particle ($m_2=m$) be initially at rest (${\overrightarrow{u}}_2=0$).
After the elastic collision, let the velocity of the heavier particle be $v_1$ at an angle $\theta$ above the horizontal, and the velocity of the lighter particle be $v_2$ at an angle $\varphi$ below the horizontal.

1. Conservation of Linear Momentum

Along the horizontal direction (x-axis):
$m_{1}u=m_1{v}_{1}cos\theta +m_2{v}_{2}cos\varphi$

Substituting $m_1=2m$ and $m_2=m$:
$2mu=2m{v}_{1}cos\theta +m{v}_{2}cos\varphi$
$2u-2v_{1}cos\theta =v_{2}cos\varphi$ --- (Equation 1)

Along the vertical direction (y-axis):
$0=m_1{v}_{1}sin\theta -m_2{v}_{2}sin\varphi$

Substituting $m_1=2m$ and $m_2=m$:
$2m{v}_{1}sin\theta =m{v}_{2}sin\varphi$
$2v_{1}sin\theta =v_{2}sin\varphi$ --- (Equation 2)

To eliminate the angle $\varphi$, we square and add Equation 1 and Equation 2:
${(2u-2v_{1}cos\theta )}^2+{(2v_{1}sin\theta )}^2=v_2^2({cos}^2\varphi +{sin}^2\varphi )$
$4u^2-8u{v}_{1}cos\theta +4v_1^2{cos}^2\theta +4v_1^2{sin}^2\theta =v_2^2$
$4u^2-8u{v}_{1}cos\theta +4v_1^2=v_2^2$ --- (Equation 3)

2. Conservation of Kinetic Energy (Perfectly Elastic Collision)

$\frac{1}{2}{m}_1{u}^2=\frac{1}{2}{m}_1{v}_1^2+\frac{1}{2}{m}_2{v}_2^2$

Substituting the mass values:
$\frac{1}{2}\left(2m\right)u^2=\frac{1}{2}\left(2m\right)v_1^2+\frac{1}{2}m{v}_2^2$
$2u^2=2v_1^2+v_2^2$
$v_2^2=2u^2-2v_1^2$ --- (Equation 4)

3. Combining Equations to Solve for the Angle

Equating the expressions for $v_2^2$ from Equation 3 and Equation 4:
$4u^2-8u{v}_{1}cos\theta +4v_1^2=2u^2-2v_1^2$
$6v_1^2-8u{v}_{1}cos\theta +2u^2=0$

Dividing the entire equation by 2 gives:
$3v_1^2-4u{v}_{1}cos\theta +u^2=0$

This is a quadratic equation in terms of $v_1$. Since the final speed $v_1$ must be a real physical quantity, the discriminant ($D=B^2-4AC$) of this quadratic equation must be greater than or equal to zero:
$D={(-4ucos\theta )}^2-4\left(3\right)\left(u^2\right)\ge 0$
$16u^2{cos}^2\theta -12u^2\ge 0$

Since $u>0$, we can divide by $4u^2$:
$4{cos}^2\theta -3\ge 0$
${cos}^2\theta \ge \frac{3}{4}$

Taking the square root for the acute scattering angle $\theta$:
$cos\theta \ge \frac{\sqrt{3}}{2}$

Since $cos\theta$ decreases as $\theta$ increases from 0 to $\pi /2$, this inequality translates directly to:
$\theta \le \frac{\pi }{6}$

Thus, the maximum possible angular deviation of the heavier particle is:
${\theta }_{max}=\frac{\pi }{6}\text{radians}$