Change in the energy of photon due to recoil of the nucleus

[Pushoam]

Homework Statement

Homework Equations

The Attempt at a Solution

Since energy of photon is very low compared to the rest mass energy of the nucleus, I consider non – relativistic calculation.

Conservation of linear momentum gives : momentum of nucleus = momentum of photon = p...(1)

Conservation of kinetic energy gives : $\frac { p^2 c^2}{2mc^2} + pc = 14.4keV$ ...(2)

Solving the above equation gives,

pc = 14.39999806 keV, m = 57 $m_p$

So, the energy of the photon gets reduced by 2m eV. Is this correct?
[/B]

 

[TSny]

Your work looks correct. Notice how your method of solving the equation requires keeping a large number of significant figures. There are ways to avoid this.

For example, let $E = pc$ and let $E_0 = 14.4$ kev. So your equation (2) is $\frac{E^2}{2mc^2} +E = E_0$. Solve this quadratic equation symbolically for $E$. Then make an approximation before plugging in numbers using the fact that $E_0 << mc^2$.

Or, rearrange (2) to get $E_0 - E = \frac{E^2}{2mc^2}$. Since $E$ will be close to $E_0$, you can let $E = E_0$ on the right hand side to obtain a good approximation for $E_0 - E$.

 

[Pushoam]

Your work looks correct. Notice how your method of solving the equation requires keeping a large number of significant figures. There are ways to avoid this.

For example, let $E = pc$ and let $E_0 = 14.4$ kev. So your equation (2) is $\frac{E^2}{2mc^2} +E = E_0$. Solve this quadratic equation symbolically for $E$. Then make an approximation before plugging in numbers using the fact that $E_0 << mc^2$.

Or, rearrange (2) to get $E_0 - E = \frac{E^2}{2mc^2}$. Since $E$ will be close to $E_0$, you can let $E = E_0$ on the right hand side to obtain a good approximation for $E_0 - E$.

Thanks for the insights. It's easy and clear.