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When a silver foil (z=47) was used in an α-ray scattering experiment, the number of α- particles scattered at ${30^\circ }$ was found to be 200 per minute. If the silver foil is replaced by aluminum (Z=13) foil of same thickness, the number of α particles scattered per minute at ${30^\circ }$ is nearly equal to :
A) 15
B) 30
C) 10
D) 20

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Last updated date: 20th Jun 2024
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Answer
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Hint: In this question the alpha particle is being scattered with silver and aluminum foil at an angle \[\theta = 30^\circ \]. We have to find the number of particles scattered by aluminum foil. For this we are going to use following formula of number of particles,
\[N = \dfrac{{{{\left( {Ze} \right)}^2}}}{{{{\left( {\sin \dfrac{\theta }{2}} \right)}^4}}}\]
Where,
N is the number of particles scattered
Z is the atomic number of material of foil
e is the charge of electron
$\theta $ is the angle of scattering

Complete step by step solution:
Given-
Number of alpha particles scattered with silver foil \[{N_S} = 200\]
Angle of scattering \[\theta = 30^\circ \]
Let ${N_S}$ is the number of alpha particles scattered by Silver foil and ${N_{Al}}$ is the number of alpha particles scattered by Aluminum foil.
Let \[e\] is the charge on the electron and $\theta $ is the angle of scattering.
We are going to use the following formula to find the number of scattered alpha particles.
Number of scattered particles by silver foil,
\[\Rightarrow {N_S} = \dfrac{{{{\left( {{Z_S}e} \right)}^2}}}{{{{\left( {\sin \dfrac{\theta }{2}} \right)}^4}}}\]……….. (1)
Number of scattered particles by Aluminum foil,
\[\Rightarrow {N_{Al}} = \dfrac{{{{\left( {{Z_{Al}}e} \right)}^2}}}{{{{\left( {\sin \dfrac{\theta }{2}} \right)}^4}}}\]………… (2)
Since $\theta $ is same in both the cases
Now dividing equation (1) and (2)
\[\Rightarrow \dfrac{{{N_S}}}{{{N_{Al}}}} = \dfrac{{{{\left( {{Z_S}e} \right)}^2}}}{{{{\left( {{Z_{Al}}e} \right)}^2}}}\]
\[\Rightarrow \dfrac{{{N_S}}}{{{N_{Al}}}} = \dfrac{{{{\left( {{Z_S}} \right)}^2}}}{{{{\left( {{Z_{Al}}} \right)}^2}}}\]
Putting the values of \[{Z_S}\]and \[{Z_{Al}}\]
\[\Rightarrow \dfrac{{{N_S}}}{{{N_{Al}}}} = \dfrac{{{{\left( {47} \right)}^2}}}{{{{\left( {13} \right)}^2}}}\]
\[\Rightarrow {N_{Al}} = \dfrac{{{{\left( {13} \right)}^2}}}{{{{\left( {47} \right)}^2}}}{N_S}\]
Putting the value of \[{N_S} = 200\]
\[\Rightarrow {N_{Al}} = \dfrac{{{{\left( {13} \right)}^2}}}{{{{\left( {47} \right)}^2}}} \times 200\]
\[\Rightarrow {N_{Al}} = 15.3\]
The number of alpha particles scattered by Aluminum foil is \[{N_{Al}} = 15.3\].

The number of alpha particles scattered by Aluminum foil is \[{N_{Al}} \simeq 15\].

Note: In this question scattering of alpha particles is taking place. Scattering of any particle is the deflection of a particle from its original path. Scattering happens in rays of light also. Scattering is a very important phenomenon in Physics. We see so many examples of scattering around us, for example the blue colour of sky and the colour of sun appear red at the time of sunrise and sunset. These are two very common examples of scattering.