In the Rutherford scattering experiment, the number of α-particles scattered at \[{60^ \circ }\] is \[5 \times {10^6}\] . The number of \[\alpha - \] particles scattered at \[{120^ \circ }\] will be
A. \[15 \times {10^6} \\ \]
B. $\dfrac{3}{5} \times {10^6} \\ $
C. $\dfrac{5}{9} \times {10^6}$
D. None of these

Hint:Rutherford discovered that a small proportion of particles had 180 degree deflections. Such deflections could only be explained by supposing that the positive charge and nearly all of the atom's mass were concentrated at the foil's atoms' centre. To find the number of \[\alpha - \] particles scattered at \[{120^ \circ }\] we will use the inverse proportionality between the number of particles and the scattering angle.

Formula Used:
Number of \[\alpha - \] particles,
$N \propto \dfrac{1}{{{{\sin }^4}\dfrac{\theta }{2}}}$
where $\theta $ is the scattering angle of \[\alpha - \] particles after repulsion from the nucleus of the atom.

Complete step by step solution:
Given: Number of \[\alpha - \] particles scattered at \[{60^ \circ }\] angle is, ${N_{{{60}^ \circ }}} = 5 \times {10^6}$
Let ${N_{{{120}^ \circ }}}$ be the number of \[\alpha - \] particles scattered at \[{120^ \circ }\] angle. Now, we know that,
$N \propto \dfrac{1}{{{{\sin }^4}\dfrac{\theta }{2}}}$
This implies that,
${N_{{{60}^ \circ }}} \propto \dfrac{1}{{{{\sin }^4}\dfrac{{60}}{2}}}$.........(1)
And,
\[{N_{{{120}^ \circ }}} \propto \dfrac{1}{{{{\sin }^4}\dfrac{{120}}{2}}}\]..........(2)

Dividing equations (1) and (2) we get,
\[\dfrac{{{N_{{{120}^ \circ }}}}}{{{N_{{{60}^ \circ }}}}} = \dfrac{{\dfrac{1}{{{{\sin }^4}\dfrac{{120}}{2}}}}}{{\dfrac{1}{{{{\sin }^4}\dfrac{{60}}{2}}}}} \\ \]
Simplifying this,
\[\dfrac{{{N_{{{120}^ \circ }}}}}{{{N_{{{60}^ \circ }}}}} = \dfrac{{{{\sin }^4}\dfrac{{60}}{2}}}{{{{\sin }^4}\dfrac{{120}}{2}}} \\ \]
Further simplifying,
\[\dfrac{{{N_{{{120}^ \circ }}}}}{{{N_{{{60}^ \circ }}}}} = \dfrac{{{{\sin }^4}{{30}^ \circ }}}{{{{\sin }^4}{{60}^ \circ }}} \\ \]
Substituting the known values, we get,
\[{N_{{{120}^ \circ }}} = \dfrac{{{{\left( {\dfrac{1}{2}} \right)}^4}}}{{{{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^4}}}5 \times {10^6} \\ \]
Thus,
\[\therefore {N_{{{120}^ \circ }}} = \dfrac{5}{9} \times {10^6}\]

Hence, option C is the answer.

Note: While solving this question remember that $\theta $ is not the angle with which the \[\alpha - \] particles are bombarded. Rather it is the angle with which the \[\alpha - \] particles are scattered after repulsion from the nucleus. Hence, not all particles are scattered with the same angle. The particles nearest to the nucleus will bounce back whereas other particles get deflected by different angles according to their distance from the nucleus.