When U238 nucleus originally at rest, decays by emitting an alpha particle having a speed u, the recoil speed of the residual nucleus is

  • Option 1)

    \frac{4u}{238}\;

  • Option 2)

    \; -\frac{4u}{234}\;

  • Option 3)

    \; \frac{4u}{234}\;

  • Option 4)

    \; -\frac{4u}{238}

 

Answers (1)

As we learnt in

α -decay -

^{A}_{Z}X\rightarrow ^{A-4}_{Z-2}Y+^{4}_{2}He+Q

 

- wherein

Q\: value = \left ( M_{X} -M_{Y}-M_{He}\right )c^{2}

 

 

From momentum conservation 

Pi = Pf 

Pi = 0 = Pf 

\Rightarrow\ \; 4.u+234.V=0

\therefore\ \;V=-\frac{4u}{234}

Correct option is 2.


Option 1)

\frac{4u}{238}\;

This is an incorrect option.

Option 2)

\; -\frac{4u}{234}\;

This is the correct option.

Option 3)

\; \frac{4u}{234}\;

This is an incorrect option.

Option 4)

\; -\frac{4u}{238}

This is an incorrect option.

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