Question

An atom of rest mass m0

is at rest in a laboratory and absorbs a photon of frequency ν
. Find the velocity and mass of the recoiling particle.

The answers are given in the back of the book as

u=chνhν+m0c2m=(m20+2hνm0c2)1/2.
I've found that I can figure out the velocity by starting with conservation of momentum and energy

hνc=γm0uhν+m0c2=γm0c2,
(where of course γ=[1−u2c2]−1/2
) then eliminating the relativistic mass γm0
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=c(2m0c2hν+h2ν2)1/2hν+m0c2(from conservation of energy)
u=chν(h2ν2+m20c4)1/2(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 γm0
, but quickly got a monstrosity of m0
's, h
's, ν
'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? A)
�
=
�
ℎ
�
ℎ
�
+
�
0
�
2
�
0
u=chν
m
0
​

hν+m
0
​
c
2

​

​
,
�
=
�
0
2
+
2
ℎ
�
�
0
�
2
�
4
m=
c
4

m
0
2
​
+2hνm
0
​
c
2

​

​


B)
�
=
�
ℎ
�
2
�
0
�
2
ℎ
�
+
ℎ
2
�
2
ℎ
�
+
�
0
�
2
u=chν
hν+m
0
​
c
2

2m
0
​
c
2
hν+h
2
ν
2

​

​
,
�
=
�
0
1
−
�
2
�
2
m=
1−
c
2

u
2

​

​

m
0
​

​


C)
�
=
�
ℎ
�
ℎ
2
�
2
+
�
0
2
�
4
ℎ
�
+
�
0
�
2
u=chν
hν+m
0
​
c
2

h
2
ν
2
+m
0
2
​
c
4

​

​
,
�
=
�
0
2
+
2
ℎ
�
�
0
�
2
�
4
m=
c
4

m
0
2
​
+2hνm
0
​
c
2

​

​


D)
�
=
�
ℎ
�
ℎ
2
�
2
+
�
0
2
�
4
ℎ
2
�
2
+
2
ℎ
�
�
0
�
2
u=chν
h
2
ν
2
+2hνm
0
​
c
2

h
2
ν
2
+m
0
2
​
c
4

​

​
,
�
=
�
0
1
−
�
2
�
2
m=
1−
c
2

u
2

​

​

m
0
​

​

Answer 1

To find the velocity and mass of the recoiling particle after absorbing a photon, you should start with both energy and momentum conservation. Then, eliminate the relativistic mass γm0 and solve for the velocity. Finally, substitute the velocity into the relativistic mass γm0 to find the mass of the recoiling particle.

Step-by-step explanation:

The correct way to approach this problem is to start with both energy and momentum conservation. By considering both conservation laws, you can eliminate the relativistic mass γm0 and solve for the velocity u. The correct expression for the velocity is u=c(2m0c2hν+h2ν2)1/2 / (hν+m0c2).

As for the mass, after finding the velocity u, you can substitute it into the relativistic mass γm0 to find the mass of the recoiling particle. The correct expression for the mass is m=(m02+2hνm0c2)1/2.