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)

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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