Question

In Rutherford’s experiment the number of alpha particles scattered through an angle of \[{60^0}\] by a silver foil is 200 per minute. When the silver foil is replaced by a copper coil of the same thickness, the number of a particles scattered through an angle of \[{60^0}\] per minute is -

Answer 1

Hint:In this question we use the concept of Rutherford atomic experiment. In this experiment Rutherford determined the scattering formula which gives the relation between the number of particles scattered and scattering angle.

Formula used:
Rutherford scattering formula is given as,
\[N \propto \dfrac{{{Z^2}}}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}\]
Where \[N(\theta )\]is the number of alpha particles scattered.
Z is the atomic number of the foil element used.
\[\theta \] is the scattering angle.
K is the constant.

Complete step by step solution:
As we know that the number of alpha particles scattered by a foil is given by Rutherford scattering formula,
\[N \propto \dfrac{{{Z^2}}}{{{{\sin }^4}\left( {\dfrac{\theta }{2}} \right)}}\]
Substituting the given values, we get
For the silver foil (Z=47),
\[{N_S} \propto \dfrac{{{{(47)}^2}}}{{{{\sin }^4}\left( {\dfrac{{{{60}^0}}}{2}} \right)}} \\ \]
\[\Rightarrow 200 \propto \dfrac{{{{(47)}^2}}}{{{{\sin }^4}\left( {\dfrac{{{{60}^0}}}{2}} \right)}}\] …...1

For the copper foil (Z=29),
\[{N_C} \propto \dfrac{{{{(29)}^2}}}{{{{\sin }^4}\left( {\dfrac{{{{60}^0}}}{2}} \right)}}\] …...2
From equation 1 and 2, we have
\[\dfrac{{200}}{{{N_C}}} = \dfrac{{{{(47)}^2}}}{{{{(29)}^2}}} \\ \]
\[\Rightarrow {N_C} = 200 \times \dfrac{{{{(29)}^2}}}{{{{(47)}^2}}} = 525.34 \\ \]
\[\therefore {N_C} \approx 525\]
The number of alpha particles scattered per min for copper foil (\[{N_C} = 525.34\]) is 525.

Therefore, when the silver foil is replaced by a copper coil of the same thickness, the number of particles scattered through an angle of \[{60^0}\] per minute is 525.

Note: As Rutherford did the gold foil experiment to determine the nature of an atom. In Rutherford’s experiment it is concluded that there is a dense centre inside an atom known as the nucleus which is positively charged. Rutherford in his experiment found that the number of alpha particles scattered per unit time is directly proportional to the atomic number of the foil element.