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QJEEAdv19Ph2cbc: JEE Advanced 2019 Physics - Class 12 JEEAdv19Ph2cbc Solution: Suppose a nucleus at rest and in ground state undergoes -decay…
Suppose a nucleus at rest and in ground state undergoes -decay to a nucleus in its excited state. The kinetic energy of the emitted particle is found to be 4.44 MeV. nucleus then goes to its ground state by -decay. The energy of the emitted photon is ............ keV.
[Given : atomic mass of = 226.005 u, atomic of = 222.000 u, atomic mass of particle = 4.000 u, 1 u = 931 MeV/e2, c is speed of the light]
[Given : atomic mass of = 226.005 u, atomic of = 222.000 u, atomic mass of particle = 4.000 u, 1 u = 931 MeV/e2, c is speed of the light]
Step-by-Step Verified Solution
Given, atomic mass of ,
atomic mass of .
atomic mass of particle
where, c = speed of the light
Total energy emitted
Kinetic energy of
Kinetic energy of Rn = = 0.08 MeV
atomic mass of .
atomic mass of particle
where, c = speed of the light
Total energy emitted
Kinetic energy of
Kinetic energy of Rn = = 0.08 MeV