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Atoms and Nuclei question

2022 · 29 Jun · Shift 1 · Q64
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Atoms and Nuclei question

2022 · 29 Jun · Shift 1 · Q64

JEE MainPhysicsAtoms and NucleiNumerical+4 / −1
d1\sqrt {{d_1}}d1​​ and d2\sqrt {{d_2}}d2​​ are the impact parameters corresponding to scattering angles 60 ∘^\circ∘ and 90 ∘^\circ∘ respectively, when an α\alphaα particle is approaching a gold nucleus. For d1 = x d2, the value of x will be ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 3

  1. Use Rutherford scattering relation

For scattering of an \alpha-particle by a nucleus, the impact parameter bbb and scattering angle θ\thetaθ are related by

b=Ccot⁡(θ2)b = C \cot\left(\frac{\theta}{2}\right)b=Ccot(2θ​)

where CCC is a constant for the given projectile and target.

Here, the problem states that the impact parameters are d1\sqrt{d_1}d1​​ and d2\sqrt{d_2}d2​​ for scattering angles 60∘60^\circ60∘ and 90∘90^\circ90∘ respectively.

So,

d1=Ccot⁡(60∘2)=Ccot⁡30∘\sqrt{d_1} = C \cot\left(\frac{60^\circ}{2}\right)= C\cot 30^\circd1​​=Ccot(260∘​)=Ccot30∘ d2=Ccot⁡(90∘2)=Ccot⁡45∘\sqrt{d_2} = C \cot\left(\frac{90^\circ}{2}\right)= C\cot 45^\circd2​​=Ccot(290∘​)=Ccot45∘
  1. Evaluate cot values
cot⁡30∘=3,cot⁡45∘=1\cot 30^\circ = \sqrt{3}, \qquad \cot 45^\circ = 1cot30∘=3​,cot45∘=1

Thus,

d1=C3,d2=C\sqrt{d_1} = C\sqrt{3}, \qquad \sqrt{d_2} = Cd1​​=C3​,d2​​=C
  1. Square both relations
d1=3C2,d2=C2d_1 = 3C^2, \qquad d_2 = C^2d1​=3C2,d2​=C2

Hence,

d1=3d2d_1 = 3d_2d1​=3d2​

Comparing with

d1=xd2d_1 = x d_2d1​=xd2​

we get

x=3x=3x=3
  1. Comparison with stored answer

Stored correct answer = 333

Our derived answer = 333

So they agree.

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