Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are λ₁ and λ₂, their de Broglie wavelength in the frame of reference attached to their center of masses: (a) λCM = λ₁ = λ₂ (b) λCM = (2λ₁λ₂)/(√(λ₁²+λ₂²)) (c) 1λCM = (1)/(λ₁) + (1)/(λ₂) (d) λCM = ((λ₁+λ₂)/(2))

Answer: (b). Equal masses: vCM = ( v₁ + v₂)/(2), so each electron's velocity relative to CM is ( v₁ - v₂)/(2) in magnitude. v₁ ⊥ v₂: pCM = 1/2√(p₁² + p₂²) = h/2√((1)/(λ₁²) + (1)/(λ₂²)) λCM = hpCM = (2λ₁λ₂)/(√(λ₁² + λ₂²)) So the answer is (b).

Two electrons are moving with non-relativistic speed perpendicular to… | JEE Main 2018 (15 Apr, Shift 1)

Q130JEE Main 2018 · 15 APR · SHIFT 1Difficult

Matter Waves: de Broglie Wavelength

Two electrons are moving with non-relativistic speed perpendicular to each other. If corresponding de Broglie wavelength are $\lambda_1$ and $\lambda_2,$ their de Broglie wavelength in the frame of reference attached to their center of masses:

Show answer and solution

Answer: (b)

Equal masses: $\vec v_{CM} = \dfrac{\vec v_1 + \vec v_2}{2}$ , so each electron's velocity relative to CM is $\dfrac{\vec v_1 - \vec v_2}{2}$ in magnitude.
$\vec v_1 \perp \vec v_2$ : $p_{CM} = \dfrac{1}{2}\sqrt{p_1^2 + p_2^2} = \dfrac{h}{2}\sqrt{\dfrac{1}{\lambda_1^2} + \dfrac{1}{\lambda_2^2}}$
$\lambda_{CM} = \dfrac{h}{p_{CM}} = \dfrac{2\lambda_1\lambda_2}{\sqrt{\lambda_1^2 + \lambda_2^2}}$
So the answer is (b).

Report a mistake

Matter Waves: de Broglie Wavelength question from JEE Main 2018 (15 Apr, Shift 1)

More questions on Matter Waves: de Broglie Wavelength