Question

In the Rutherford experiment, the number of alpha particles scattered at an angle of $ {{\theta = 6}}{{{0}}^{{0}}} $ is 36 per minute. How many alpha particles are scattered per minute at an angle of $ {{\theta = 4}}{{{5}}^{{0}}} $ $ \left[ {\sin {{22.5}^0}} \right] = 0.3827 $
(A) 210
(B) 105
(C) 72
(D) 144

Answer 1

Hint: The Rutherford’s experiment, known as the “alpha particle-scattering experiment” and was performed to prove that the position of the nucleus is at the centre of the atom and that the mass of the nucleus only determines the mass of the atom as the electrons are massless.
 $ {{N}}\left( {{\theta }} \right){{ = }}\dfrac{{{{{Z}}^{{2}}}}}{{{{si}}{{{n}}^{{4}}}\left( {\dfrac{{{\theta }}}{{{2}}}} \right)}} $ ;
Where, $ {{N}}\left( {{\theta }} \right) $ is the number of alpha particles scattered, Z is the atomic number of the element, $ {{\theta }} $ is the scattering angle.

Complete Stepwise Solution:
According to the question, when $ {{N}}\left( {{\theta }} \right) $ = 36 particles per minute, $ {{\theta = 6}}{{{0}}^{{0}}} $ , we can get the value of Z.
 $ {{36 = }}\dfrac{{{{{Z}}^{{2}}}}}{{{{si}}{{{n}}^{{4}}}\left( {\dfrac{{60}}{{{2}}}} \right)}} = \dfrac{{{{{Z}}^{{2}}}}}{{{{si}}{{{n}}^{{4}}}\left( {30} \right)}} $
Rearranging:
Or, $ {{ }}{{{Z}}^{{2}}}{{ = 36 si}}{{{n}}^{{4}}}\left( {30} \right) = \dfrac{{36}}{{\dfrac{1}{{16}}}} = 36 \times 16 $
For, $ {{\theta = 4}}{{{5}}^{{0}}} $ , $ {{ }}{{{Z}}^{{2}}}{{ = }}36 \times 16 $ , putting the values in the equation we get, $ {{N}}\left( {{\theta }} \right){{ = }}\dfrac{{36 \times 16}}{{{{si}}{{{n}}^{{4}}}\left( {\dfrac{{45}}{{{2}}}} \right)}} = 36 \times 16 \times {\left( {\dfrac{1}{{0.3827}}} \right)^4} = 104.89 \simeq 105 $
So the number of particles emitted is equal to 105, option B is the correct answer.

Note:
The experiment was performed by using a very thin gold foil which was placed at the centre of a circular shaped detector that was coated with zinc sulphide. A certain amount of radium was kept in a box made of lead and this was used as the source of radioactive radiation. The alpha radiation this emitted by radium hit the gold foil and most of the radiations passed through the gold foil and probably 1 out of 20,000 alpha particles would suffer head-one deflection after colliding with the nucleus present at the centre of the gold atom getting scattered by $ {{\theta = 18}}{{{0}}^{{0}}} $ almost. This experiment pointed out to two things:
Firstly, the nucleus is made up of electrons which have negligible mass and secondly, the nucleus of the atom lies at the centre of the atom and bears positive charge and all the mass of the atom is due to this nucleus.