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

[sadhu]

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 ??

 

[chbeatbama]

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.

 

[sir-pinski]

The solution is correct, but there are a few errors in the notation and units.

Firstly, the unit for energy is usually measured in joules (J) or electron volts (eV), so it would be more appropriate to use these units instead of just "e". Additionally, in the momentum conservation equation, the units do not match up. The left side is in units of mass times velocity (kg*m/s) while the right side is in units of energy divided by the speed of light (J*s/m). It would be more accurate to write it as Mv = -e/c.

In the second equation, the units also do not match up. The left side is in units of mass times velocity squared (kg*m^2/s^2) while the right side is in units of energy (J/eV). It would be more accurate to write it as Mv^2/2 + e = E.

Additionally, it would be helpful to define the variables used in the equations. For example, it is not clear what "E" represents in this context. Is it the total energy of the atom or the energy of the photon emitted? Clarifying this would make the solution easier to understand.

Overall, the solution is correct but it would benefit from clearer notation and units. It is important to pay attention to these details in order to avoid errors and confusion.