Suppose a 88226 Ra nucleus at rest and in ground state undergoes α-decay to a 86222 Rn nucleus in its excited state. The kinetic energy of the emitted α particle is found to be 4.44 MeV. 86222 Rn nucleus then goes to its ground state by γ-decay. The energy of the emitted γ photon is ____ keV.[Given : atomic mass of 88226 Ra = 226.005 u, atomic mass of 86222 Rn = 222.000 u, atomic mass of α particle = 4.000 u, 1 u = 931 MeV/c2, c is speed of the light]
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answer is 135.00.
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Detailed Solution
88226Ra⇒86222Rn+24HeΔm=mRa−mRn+ma=226.005−(222+4)=0.005uQ - Value =5×10−3×931MeV=4.655MeV, Since, KE∝1m⇒ K.E. of 86222Rn=4.44×4222=0.08MeV Energy of γ=(4.655−4.44−0.08)MeV=135keV