
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
Answer
162.3k+ views
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.
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.
Recently Updated Pages
JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

Young's Double Slit Experiment Step by Step Derivation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

JEE Energetics Important Concepts and Tips for Exam Preparation

JEE Isolation, Preparation and Properties of Non-metals Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric field due to uniformly charged sphere class 12 physics JEE_Main

Degree of Dissociation and Its Formula With Solved Example for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Charging and Discharging of Capacitor

Wheatstone Bridge for JEE Main Physics 2025

Formula for number of images formed by two plane mirrors class 12 physics JEE_Main

In which of the following forms the energy is stored class 12 physics JEE_Main
