Question 657: Atoms & Nuclei

Question

A heavy nucleus N, at rest, undergoes fission N $$	o$$ P + Q, where P and Q are two lighter nuclei. Let $$\delta$$ = 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 speeds 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?

Accepted Answer

(p
total
) = 0 $$\Rightarrow$$ (p
total
)
f
= 0
So, momentum of one nucleus is p in forward direction, then momentum of the other nucleus will be p in backwards direction.
Energy released = ($$\Delta$$m)c
2
= $$\delta$$c
2
= KE
P
+ KE
Q
$$K{E_P} = {{{p^2}} \over {2{m_P}}},K{E_Q} = {{{p^2}} \over {2{m_Q}}}$$
$$ \Rightarrow K{E_P}:K{E_Q} = {1 \over {{m_P}}}:{1 \over {{m_Q}}} = {m_Q}:{m_P}$$
$$K{E_P} = \left( {{{{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$$
$$K{E_Q} = \left( {{{{m_P}} \over {{m_P} + {m_Q}}}} \right)\delta {c^2}$$
$$K{E_P} + K{E_Q} = \delta {c^2}$$
$${{{p^2}} \over {2{m_P}}} + {{{p^2}} \over {2{m_Q}}} = \delta {c^2} \Rightarrow p = c\sqrt {2\left( {{{{m_P}{m_Q}} \over {{m_P} + {m_Q}}}} \right)\delta } $$

Answer

$${E_P} + {E_Q} = {c^2}\delta $$

Answer

$${E_P} = \left( {{{{M_P}} \over {{M_P} + {M_Q}}}} \right){c^2}\delta $$

Answer

$${{{v_P}} \over {{v_Q}}} = {{{M_Q}} \over {{M_P}}}$$

Answer

The magnitude of momentum for P as well Q is $$c\sqrt {2\mu \delta } $$, where $$\mu  = {{{M_P}{M_Q}} \over {({M_P} + {M_Q})}}$$

Atoms & Nuclei

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