PhysicsCBSE Class 12 · Sample paper, session 2024-25 · 1 mark · MCQ

CBSE Class 12 Physics sample paper 2024-25: question 8

From the official CBSE sample question paper for 2024-25, section A, question 8, for 1 mark. Try it first; the hints and our step-by-step solution are below.

Q8 · 1 mark · MCQAtoms

The distance of closest approach of an alpha particle is $d$ when it moves with a speed V towards a nucleus. Another alpha particle is projected with higher energy such that the new distance of the closest approach is $d /2$ . What is the speed of projection of the alpha particle in this case?

(A)V/2
(B) $2$ V
(C)2 V
(D)4 V

Source: official sample paper from cbseacademic.nic.in (PDF). See it in the full Sample paper 2024-25.

Hints

Open one hint at a time and try again before the next.

Hint 1

At closest approach, all KE has become electrostatic PE.

Hint 2

How does d depend on V?

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

At closest approach, kinetic energy equals potential energy: $\frac{1}{2} m V^{2} = \frac{k Z e ( 2 e )}{d}$ , so $d \propto \frac{1}{V ^{2}}$ .
½ mark
Halving d doubles $V^{2}$ , so $V^{'} = 2 V$ . The answer is (B).
½ mark

Answer: $2 V$ , option (B)

Where the marks are

$d \propto 1/ V^{2}$ (0.5 mark)
$V^{'} = 2 V$ (0.5 mark)

Revise the chapter

Chapter: Atoms (notes, formulas, where marks are lost and every past question)

Practise this chapterBoard-style questions from Atoms, one at a time, marked against the answer key and marking points.Back to the full paperSample paper 2024-25: every question, by section.

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CBSE Class 12 Physics sample paper 2024-25: question 8

Hints

Step-by-step solution

Revise the chapter

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