I'm working a problem out of d'Inverno's "Introducing Einstein's Relativity", and I'm hitting a funny issue with my algebra. The problem states:
An atom of rest mass $m_0$ is at rest in a laboratory and absorbs a photon of frequency $\nu$. Find the velocity and mass of the recoiling particle.
The answers are given in the back of the book as
$$
u=\frac{ch\nu}{h\nu + m_0c^2} \qquad m=\left(m_0^2 + \frac{2h\nu m_0}{c^2}\right)^{1/2}.
$$
I've found that I can figure out the velocity by starting with conservation of momentum and energy
$$
\frac{h\nu}{c} = \gamma m_0u \qquad h\nu + m_0c^2 = \gamma m_0c^2,
$$
(where of course $\gamma=\left[1-\frac{u^2}{c^2}\right]^{-1/2}$) then eliminating the relativistic mass $\gamma m_0$ and solving for $u$. Strangely though, in my first couple of attempts I started with conservation of only energy or momentum, and obtained answers close to, but not quite the same as, the above:
$$
u=\frac{c(2m_0c^2h\nu+h^2\nu^2)^{1/2}}{h\nu + m_0c^2} \qquad (\textrm{from conservation of energy})
$$
$$
u=\frac{ch\nu}{(h^2\nu^2+m_0^2c^4)^{1/2}} \qquad (\textrm{from conservation of momentum})
$$
Is there simply some mistake in my algebra that I haven't managed to suss out, or is the error in assuming that I can proceed from only one conservation law?
As for the mass, I attempted to simply substitute the velocity $u$ into the relativistic mass $\gamma m_0$, but quickly got a monstrosity of $m_0$'s, $h$'s, $\nu$'s, and $c$'s that bore no resemblance to the answer. Is this the correct way to approach this part of the problem, or should I start from some other relation?
EDIT:
Sofia answered the second half of my question regarding the mass of the atom. However, I'm still curious about the first part. That is, why do I obtain different results for the velocity with different equations? One approach may be more difficult or roundabout than another, but if they're all based on the same physical principles, I feel that I should get the same result regardless of which relation I begin with.
Best Answer
The solution in the book doesn't consider the recoil velocity $u$ as relativistic. So, no need of $\gamma$. But they consider simply that the mass of the atom increases, instead of $m_0$ to $m$. So, this is what they do, I mean, their equations of conservation:
(1) $h \nu /c = mu, \ \ \ $ linear momentum conservation,
(2) $(m_0c^2 + h\nu)^2 = m^2c^4 + m^2u^2c^2, \ \ \ $ energy conservation and Klein-Gordon equation.
Now, from eq. (1) I obtain
$u = \frac {h \nu}{mc}$,
and I introduce in eq. (2).
$(m_0c^2 + h\nu)^2 = m^2c^4 + h^2 \nu ^2$.
From the last equation we can extract $m$,
$m_0^2 + \frac {2h\nu m_0}{c^2} = m^2$.
The quantity on the LHS can be eventually completed to a square
$m_0^2 + \frac {2h\nu m_0}{c^2} + \frac {(h \nu)^2}{c^4} = m^2$.