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A nucleus \( \mathrm{X} \), initially at rest, undergoes alpha-decay according to the equation.
\[
{ }_{92}^{\mathrm{A}} \mathrm{X} \rightarrow+{ }_{\mathrm{Z}}^{228} \mathrm{Y}+\alpha
\]
The alpha particle produced in the above process is found to move in a circular track of radius \( 0.11 \mathrm{~m} \) in a uniform magnetic field of 3 tesla. Find the energy (in \( \mathrm{MeV} \) ) released during the process
Given that \( m(Y)=228.03 \mathrm{u} ; \mathrm{m}\left({ }_{0}^{1} \mathrm{n}\right)=1.009 \mathrm{u} \).
\[
\mathrm{m}_{2}^{4} \mathrm{He}=4.003 \mathrm{u} ; \mathrm{m}\left({ }_{1}^{1} \mathrm{H}\right)=1.008 \mathrm{u} \text {. }
\]
\( \mathrm{P} \)
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