Particle A of mass mA = m/2 moving along the x-axis with velocity v0… | JEE Main 2020 (4 Sep, Shift 1)

Q119JEE Main 2020 · 4 SEP · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength

Particle A of mass mA = $\frac{m}{2}$ moving along the x-axis with velocity v0 collides elastically with another particle B at rest having mass mB = $\frac{m}{3}$ . If both particles move along the x-axis after the collision, the change $\Delta\lambda$ in de-Broglie wavlength of particle A, in terms of its de-Broglie wavelength ( $\lambda$ 0) before collision is:

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Answer: (d)

Elastic, B at rest: $v_A' = \dfrac{m_A - m_B}{m_A + m_B}v_0 = \dfrac{m/2 - m/3}{m/2 + m/3}v_0 = \dfrac{v_0}{5}$
$\lambda_A' = \dfrac{h}{m_A v_0/5} = 5\lambda_0$
$\Delta\lambda = 5\lambda_0 - \lambda_0 = 4\lambda_0$
So the answer is (d).

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