JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Modern Physics

A nucleus A, with a finite de-broglie wavelength

λ_{A}

, undergoes spontaneous fission into two nuclei B and C of equal mass. B flies in the same direction as that of A, while C flies in the opposite direction with a velocity equal to half of that of B. The de-Broglie wavelength

λ_{B}

and

λ_{C}

of B and C are respectively:

(A) $λ_{A}, 2 λ_{A}$
(B) $2 λ_{A}, λ_{A}$
(C) $λ_{A}, \frac{λ _{A}}{2}$
(D) $\frac{λ _{A}}{2}, λ_{A}$ ✓ Correct

Answer: (D)

Solution

Take A to have mass

2 m

and speed

v_{0}

; after fission B and C each have mass

m

. Let B move forward with speed

v

and C backward with speed

v /2

. Momentum conservation gives

2 m v_{0} = m v - m \frac{v}{2} = \frac{m v}{2}

, so

v = 4 v_{0}

. The momenta are then

p_{A} = 2 m v_{0}

p_{B} = m (4 v_{0}) = 4 m v_{0}

, and

p_{C} = m (2 v_{0}) = 2 m v_{0}

. Since

λ = \frac{h}{p}

, more momentum means a shorter wavelength:

λ_{B} = \frac{h}{4 m v _{0}} = \frac{λ _{A}}{2}

, while

λ_{C} = \frac{h}{2 m v _{0}} = λ_{A}

. So B has half the wavelength of A, and C has the same wavelength as A.

JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Single correct option

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JEE Main · 2019 · 08 Apr 2019 · Shift 2 · Modern Physics

Solution

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