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Q. A nucleus with mass number $224$ initially at rest emits an $\alpha $ particle. If the $Q$ value of the reaction is $8MeV$ and only one photon of energy $2.00MeV$ is subsequently emitted after the emission of the $\alpha $ particle, the kinetic energy (in $MeV$ ) of the $\alpha $ particle is : (Approximate the answer to the nearest integer.)

NTA AbhyasNTA Abhyas 2022

Solution:

$T_{\alpha }=\frac{\left(Q - \Delta E\right) Y}{Y + 4}=\frac{\left(8 - 2\right) 220}{224}=5.89MeV$