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JEE Main 20208 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
8 January
Shift
Morning Shift
Subject
Physics
Chapter
Atomic Physics

JEE Main 2020 · 8 January · Morning Shift · Physics · Atomic Physics

The graph which depicts the results of Rutherford gold foil...

The graph which depicts the results of Rutherford gold foil experiment with α\alpha -particles is:
θ:\theta: Scattering angle
Y:Y: Number of scattered α\alpha -particles detected
(Plots are schematic and not to scale)The graph which depicts the results of Rutherford gold foil experiment with α -particles is: θ: Scattering angle Y:The graph which depicts the results of Rutherford gold foil experiment with α -particles is: θ: Scattering angle Y:The graph which depicts the results of Rutherford gold foil experiment with α -particles is: θ: Scattering angle Y:The graph which depicts the results of Rutherford gold foil experiment with α -particles is: θ: Scattering angle Y:

Options

  1. ALight is incident from glass ( μ = 1.50 ) to water ( μ = 1.33 ). Determine the range of the angle of deviation for
  2. BLight is incident from glass ( μ = 1.50 ) to water ( μ = 1.33 ). Determine the range of the angle of deviation for
  3. CLight is incident from glass ( μ = 1.50 ) to water ( μ = 1.33 ). Determine the range of the angle of deviation for
  4. DLight is incident from glass ( μ = 1.50 ) to water ( μ = 1.33 ). Determine the range of the angle of deviation for

Correct answer

D. Light is incident from glass ( μ = 1.50 ) to water ( μ = 1.33 ). Determine the range of the angle of deviation for

Step-by-step solution

Rutherford carried out an experiment in which he bombarded a thin sheet of gold foil with α\alpha-particles and then analysed the trajectory of these particles after they collided with the gold foil.

For a detector at a specific angle with respect to the incident beam, the number of particles per unit area striking the detector is given by the Rutherford formula,

Y(θ)=NinLZ2k2e44r2(KE)2sin⁡4(θ2)Y\left(\theta\right) = \dfrac{{N}_{i}nL{Z}^{2}{k}^{2}{e}^{4}}{4{r}^{2}{\left(KE\right)}^{2}\sin^{4}\left(\dfrac{\theta}{2}\right)}

Here, Ni={N}_{i} = Number of incident alpha particles,

n=n = Atoms per unit volume in target,

L=L = Thickness of target,

Z=Z = Atomic number of target,

e=e = Electron charge,

k=k = Coulomb's constant,

r=r = Distance between target and detector,

KE=KE = Kinetic energy of alpha particles,

θ=\theta = Scattering angle.

Therefore, Y∝1sin⁡4(θ2)Y ∝ \dfrac{1}{\sin^{4}\left(\dfrac{\theta}{2}\right)}

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