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Binding Energy

A heavy nucle...

Question

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 ?

E_{P} + E_{Q} = c^{2} δ

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E_{P} = (\frac{M_{P}}{M_{P} + M_{Q}}) c^{2} δ

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\frac{v_{P}}{v_{Q}} = \frac{M_{Q}}{M_{P}}

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The magnitude of momentum for

P

as well as

Q

c \sqrt{2 μ δ}

, where

μ = \frac{M_{P} M_{Q}}{(M_{P} + M_{Q})}

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Solution

The correct option is D The magnitude of momentum for $P$ as well as $Q$ is $c \sqrt{2 μ δ}$ , where $μ = \frac{M_{P} M_{Q}}{(M_{P} + M_{Q})}$
Given:
$N \to P + Q$

Energy released in this fission will be,

$(M_{N} - M_{P} - M_{Q}) c^{2} = δ c^{2}$

This will be distributed kinetic energy of $P$ and $Q$ .

$\Rightarrow E_{P} + E_{Q} = δ c^{2} . . . (1)$

By conservation of momentum,

So, $\frac{v_{P}}{v_{Q}} = \frac{M_{Q}}{M_{P}} . . . (2)$

Kinetic energy be written as $K E = \frac{p^{2}}{2 M}$

So equation $(1)$ , became,

$\Rightarrow \frac{p^{2}}{2 M_{P}} + \frac{p^{2}}{2 M_{Q}} = δ c^{2}$

$\Rightarrow \frac{p^{2}}{2} = \frac{M_{P} M_{Q}}{M_{P} + M_{Q}} δ c^{2} . . . . . . (3)$

$∴ E_{P} = \frac{p^{2}}{2 M_{P}} = \frac{M_{Q}}{M_{P} + M_{Q}} δ c^{2}$

Similarly, $E_{Q} = \frac{M_{P}}{M_{P} + M_{Q}} δ c^{2}$

From eq. $(3)$ , momentum of $P$ or $Q$ will be,

$p = \sqrt{\frac{2 M_{P} M_{Q}}{M_{P} + M_{Q}} δ c^{2}} = c \sqrt{2 μ δ}$

Where, $μ = \frac{M_{P} M_{Q}}{M_{P} + M_{Q}}$

Hence, $(A), (C)$ and $(D)$ are the correct options

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