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

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SUMMARY

The recoil energy of an excited atom during de-excitation can be calculated using the formula K = Ephoton2/2mc2, where mc2 represents the rest mass energy of the atom. The discussion establishes that the total energy of the process is E, and through momentum conservation, the recoil velocity v is derived as v = -e/(Mc). The recoil energy is subsequently expressed in terms of the emitted photon energy e and the mass M of the atom, leading to the conclusion that K is proportional to the square of the photon energy divided by twice the product of the mass and the speed of light squared.

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sadhu
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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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