Finding the recoil energy of an excited atom when it de-excites

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The discussion focuses on calculating the recoil energy of an excited atom as it de-excites and emits a photon. It begins with the conservation of momentum and energy equations, leading to the expression for recoil velocity and energy. The recoil energy is derived from substituting the recoil velocity into the energy conservation equation. The final formula for recoil energy (K) is presented as K = E^2_photon / 2mc^2, where mc^2 represents the rest mass energy of the atom. The calculations indicate that mc^2 values for atoms and nuclei range from 10^9 to 10^11 eV.
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I know that i am commiting some serious errors in this , but just can't notice them
suppose we have to find the recoil energy of an excited atom when it de excites

suppose the total energy of the process is E

an atom will emit a photon on returning to normal state . energy of the photon is... e

recoil velocity of the atom is v;
mass of the atom is M

Mv+e/c=0....momentum conservation...c=velocity of light.initialy the atomis at rest

Mv*v/2 + e=E...energy conservation


from first equation

v=-e/(Mc)...(i)

thus recoil energy must come out by substituting v

Me*e/(2*M*M*c*c)

e*e/(2Mc*c)

substituting this in energy conservation equation

e+e*e/(2*Mc*c)=E
e+e*e/(2*Mc*c)-E=0

solving for e;
\Delta=1+2E/(M*c*c)

whats is the solution ??
 
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The recoil energy (K) is given as:

K = E^{2}_{photon}/2mc^{2}

where mc2 describes the rest mass of the particle. mc2 will be of the order 109 to 1011 eV for atoms and nuclei.
 
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