CUET · PHYSICS · PYQ PAPER 2023

The kinetic energy of an \(\alpha\)-particle incident on gold foil is reduced to half of its initial value. How does the distance of closest approach change?

A becomes one-fourth
B becomes half
C becomes twice
D remains the same

Verified Solution

Answer & Solution

Correct Answer

(C) becomes twice

Step-by-step Solution

Detailed explanation

\( r_{min} \propto \frac{1}{K} \) \( \frac{r_{min,final}}{r_{min,initial}} = \frac{K_{initial}}{K_{final}} \) \( \frac{r_{min,final}}{r_{min,initial}} = \frac{K_{initial}}{K_{initial}/2} = 2 \) The distance of closest approach becomes twice.

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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Match List - I with List - II.

List - I	List - II
(A) Resolving power of an astronomical telescope	(I) \( 1.22\lambda/D \)
(B) Limit of resolution of an astronomical telescope	(II) \( \mu \sin \theta \)
(C) Numerical aperture	(III) \( D/1.22\lambda \)
(D) Fresnel distance	(IV) \( a^2/\lambda \)

Choose the correct answer from the options given below:

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LIST I	LIST II
A. Brackett series	I. Visible region
B. Balmer series	II. Ultraviolet region
C. Paschen series	III. far infrared region
D. Lyman series	IV. infrared region