Answer
Verified
85.5k+ views
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.
\[N = \dfrac{{{{\left( {Ze} \right)}^2}}}{{{{\left( {\sin \dfrac{\theta }{2}} \right)}^4}}}\]
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.
Recently Updated Pages
Name the scale on which the destructive energy of an class 11 physics JEE_Main
Write an article on the need and importance of sports class 10 english JEE_Main
Choose the exact meaning of the given idiomphrase The class 9 english JEE_Main
Choose the one which best expresses the meaning of class 9 english JEE_Main
What does a hydrometer consist of A A cylindrical stem class 9 physics JEE_Main
A motorcyclist of mass m is to negotiate a curve of class 9 physics JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
A cylinder of 10 Lcapacity at 300 Kcontaining the Hegas class 11 chemistry JEE_Main
A scooterist sees a bus 1km ahead of him moving with class 11 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
Derive an expression for maximum speed of a car on class 11 physics JEE_Main
The process requiring the absorption of energy is A class 11 chemistry JEE_Main