A nucleus ZXA emits an α-particle with velocity v. The recoil speed of the daughter nucleus is

\(\frac{{A - 4}}{{4v}}\)

\(\frac{{4v}}{{A - 4}}\)

\(\frac{v}{4}\)

A nucleus ZXA emits an α-particle with velocity v. The recoil speed of the daughter nucleus is Correct Answer \(\frac{{4v}}{{A - 4}}\)

CONCEPT:

Linear Momentum: The product of the mass and velocity of an object is called linear momentum.

p = mv

where p is linear momentum, m is mass of the body and v is the velocity.

Law of conservation of momentum: the law of conservation of momentum states that the total momentum of an isolated system remains constant.

p_i= p_f

where pf is final linear momentum and p_i is initial linear momentum.

CALCULATION:

Given that initially, ZXA is at rest so, v_i= 0

speed of α = v

Let the α particle decay equation is

ZXA —→ Z-2YA-4 + ₂α⁴

where Y is the daughter nucleus.

and vy is the velocity of the daughter nucleus.

By Principle of momentum conservation

pi = pf

m_iv_i= m_fv_f

0 = (A-4) v_y - 4 v

Recoil speed (vy) = 4v/(A-4)

So the correct answer is option 2.

Feb 20, 2025

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