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Question

The isotope 125B having a mass 12.014 u undergoes β decay to 126C. 126C has an excited state of the nucleus (126C) at 4.041 MeV above its ground state. If 125B decay to 126C, the maximum kinetic energy of the β particle in units of MeV nearly is (integer only).

Take : m(126C)=12.0 u(1u=931.5 MeV/c2,Where c is the speed of light in vacuum)


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Solution

Given, m(125B)=12.014 um(126C)=12.0 uEexcitation=4.041 MeV

125B 126C+β+γ+Q

Q=Δmc2=[m(125B)m(126C)]c2

=[12.01412.0]c2

=0.014×931.5=13.041 MeV

Now, KEmax=QEexcitation

KEmax=13.0414.041=9 MeV

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