JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · Conservation of Momentum

A heavy nucleus

N

, at rest, undergoes fission

N \to P + Q

, where

P

and

Q

are two lighter nuclei. Let

δ = M_{N} - M_{P} - M_{Q}

, where

M_{P}, M_{Q}

and

M_{N}

are the masses of

P, Q

and

N

, respectively.

E_{P}

and

E_{Q}

are the kinetic energies of

P

and

Q

, respectively. The speed of

P

and

Q

are

v_{P}

and

v_{Q}

, respectively. If

c

is the speed of light, which of the following statement(s) is(are) correct ?

(A) $E_{P} + E_{Q} = c^{2} δ$ ✓ Correct
(B) $E_{P} = (\frac{M _{P}}{M _{P} + M _{Q}}) c^{2} δ$
(C) $\frac{v _{P}}{v _{Q}} = \frac{M _{Q}}{M _{P}}$ ✓ Correct
(D)The magnitude of momentum for $P$ as well as $Q$ is $c 2 μ δ$ , where $μ = \frac{M _{P} M _{Q}}{( M _{P} + M _{Q} )}$ ✓ Correct

Answer: (A), (C), (D)

Solution

Energy released in fission:

Q = δ c^{2}

This energy appears as kinetic energy of the products:

E_{P} + E_{Q} = δ c^{2}

So (A) is correct.

By conservation of momentum (initial momentum = 0):

M_{P} v_{P} = M_{Q} v_{Q} = p

\frac{v _{P}}{v _{Q}} = \frac{M _{Q}}{M _{P}}

So (C) is correct.

Kinetic energy in terms of momentum:

E_{P} = \frac{p ^{2}}{2 M _{P}}, E_{Q} = \frac{p ^{2}}{2 M _{Q}}

Total energy:

E_{P} + E_{Q} = \frac{p ^{2}}{2 M _{P}} + \frac{p ^{2}}{2 M _{Q}} = \frac{p ^{2}}{2} (\frac{M _{P} + M _{Q}}{M _{P} M _{Q}}) = \frac{p ^{2}}{2 μ} = δ c^{2}

where

μ = \frac{M _{P} M _{Q}}{M _{P} + M _{Q}}

is the reduced mass.

Solving for

p

p = c 2 μ δ

So (D) is correct.

For (B):

E_{P} = \frac{p ^{2}}{2 M _{P}} = \frac{2 μ δ c ^{2}}{2 M _{P}} = \frac{μ δ c ^{2}}{M _{P}} = \frac{M _{Q}}{M _{P} + M _{Q}} δ c^{2}

This shows (B) is incorrect (it has

M_{P}

in numerator instead of

M_{Q}

).

Correct answers: (A), (C), (D)

JEE Advanced · 2021 · 03 Oct 2021 · Paper 2 · One or more correct options