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A nucleus with mass number 220 initially at rest emits an \alpha particle. If the Q value of the reaction is 5.5 MeV,
the kinetic energy of the \alpha particle is: 

Option: 1

4.4 MeV


Option: 2

5.4 MeV


Option: 3

5.6 MeV


Option: 4

6.5 MeV


Answers (1)

best_answer

Here, \mathrm{A}=220, \mathrm{Q}=5.5 \, \mathrm{MeV}

The kinetic energy of the \alpha particle is 

\mathrm{KE}_\alpha=\frac{\mathrm{A}-4}{\mathrm{~A}} \mathrm{Q}=\frac{220-4}{220} \times 5.5 \mathrm{MeV}=5.4 \, \mathrm{MeV}

Posted by

Pankaj Sanodiya

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