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A nucleus A, with a finite de-broglie wavelength $\lambda _{A}$ , undergoes spontaneous fission into nuclei B and C of equal mass. B files in the same direction as that of A, while C files in the opposite direction with a velocity equal to half of that of B. the de-broglie wavelengths $\lambda _{B}$ and $\lambda _{C}$ B and C are respectively:
Option 1)
$\lambda _{A},2\lambda _{A}$
Option 2)
$2\lambda _{A},\lambda _{A}$
Option 3)
$\lambda _{A},\frac{\lambda _{A}}{2}$
Option 4)
$\frac{\lambda _{A}}{2},\lambda _{A}$

Answers (1)

best_answer

Momentum convervation

initial

Assume 'B' will move in same direction as of 'A' after fission

final

$P_{initial}=P_{final}\\ m_{A}v_{A}=m_{B}v_{B}+m_{C}v_{C}$

$mV=\frac{m}{2}v_{1}+\frac{m}{2}(\frac{-v_{1}}{2})\;\;\;\;\left ( v_{C}=\frac{v_{B}}{2} \;\; and\;\;m_{B}=m_{C}=\frac{m_{A}}{2}=\frac{m}{2} \right )$ \

So $mV=\frac{mV_1}{4}$

as $\lambda =\frac{h}{p}$

$if \ \frac{h}{\lambda_{A}}= m V \ \ \ then \ \ mv_{1}=\frac{4h}{\lambda_{A}}------(1)$

$\lambda_{B}=\frac{h}{m_{B}v_{B}}=\frac{h}{\frac{m}{2}v_{1}}=\frac{2h}{mv_{1}}=\frac{2h}{\frac{4h}{\lambda_{A}}}=\frac{\lambda_{A}}{2}$

$\lambda_{C}=\frac{h}{m_{C}v_{C}}=\frac{h}{\frac{m}{2}\frac{v_{1}}{2}}=\frac{4h}{mv_{1}}=\frac{4h}{\frac{4h}{\lambda_{A}}}=\lambda_{A}$

Option 1)

$\lambda _{A},2\lambda _{A}$

Option 2)

$2\lambda _{A},\lambda _{A}$

Option 3)

$\lambda _{A},\frac{\lambda _{A}}{2}$

Option 4)
$\frac{\lambda _{A}}{2},\lambda _{A}$

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