Having a real hard time with this one.
Thanks for the help
An electron collides elastically with a hydrogen atom that is initially at rest. Assume all the motion occurs along a straight line. What fraction of the electron's initial kinetic energy is transferred to the atom? (Take the mass of the hydrogen atom to be 1840 times the mass of an electron.)
Thanks for the help
An electron collides elastically with a hydrogen atom that is initially at rest. Assume all the motion occurs along a straight line. What fraction of the electron's initial kinetic energy is transferred to the atom? (Take the mass of the hydrogen atom to be 1840 times the mass of an electron.)
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I'll use m=mass of electron M=mass of hydrogen atom, and v0=speed of electron in M's original rest frame. p0 = m*v0 is the initial momentum. I have the electron approaching along the -x axis to get a positive *velocity* for v0.
It's easier for me to find final velocities using the center of mass frame. The c.m. is moving to the right as speed:
v_cm = v0 * m/(M + m)
In the c.m. frame, the electron has velocity
v0' = v0 - v_cm = v0(1 - m/(m + M)) = v0 * M/(M+m)
...and the hydrogen has:
v1' = -v_cm = -v0 + m/(M+m)
These will get reflected on an elastic collision in the c.m. frame:
v2' = -v0' = -v0 * M / (M + m) ... final vel. of electron in c.m. frame
v3' = -v1' = v0 * m / (M + m)
Now add v_cm to put these back into the original rest frame:
v2 = v2' + v_cm = -v0 * M/(M + m) + v0 * m/(M + m)
v2 = -v0*(M-m)/(M+m)
v3 = v3' + v_cm = v0 * m/(M + m) + v0 * m/(M + m)
v3 = v0 * 2m/(M + m)
If this is correct, the sum (m/2)v2^2 + (M/2)v3^2 should equal the original KE (m/2)v0^2. I'll let you check that. Meanwhile, the ratio you're looking for is the KE of the hydrogen afterward, divided by the original electron KE:
[(M/2) v3^2 ] / [(m/2) v0^2] = (M/m) (v3/v0)^2
= (M/m) * [2m / (M + m)]^2
= 4Mm / (M + m)^2
= 4*1*1840 / 1841^2
About 0.00217 of the original energy (0.217%) is transferred.
It's easier for me to find final velocities using the center of mass frame. The c.m. is moving to the right as speed:
v_cm = v0 * m/(M + m)
In the c.m. frame, the electron has velocity
v0' = v0 - v_cm = v0(1 - m/(m + M)) = v0 * M/(M+m)
...and the hydrogen has:
v1' = -v_cm = -v0 + m/(M+m)
These will get reflected on an elastic collision in the c.m. frame:
v2' = -v0' = -v0 * M / (M + m) ... final vel. of electron in c.m. frame
v3' = -v1' = v0 * m / (M + m)
Now add v_cm to put these back into the original rest frame:
v2 = v2' + v_cm = -v0 * M/(M + m) + v0 * m/(M + m)
v2 = -v0*(M-m)/(M+m)
v3 = v3' + v_cm = v0 * m/(M + m) + v0 * m/(M + m)
v3 = v0 * 2m/(M + m)
If this is correct, the sum (m/2)v2^2 + (M/2)v3^2 should equal the original KE (m/2)v0^2. I'll let you check that. Meanwhile, the ratio you're looking for is the KE of the hydrogen afterward, divided by the original electron KE:
[(M/2) v3^2 ] / [(m/2) v0^2] = (M/m) (v3/v0)^2
= (M/m) * [2m / (M + m)]^2
= 4Mm / (M + m)^2
= 4*1*1840 / 1841^2
About 0.00217 of the original energy (0.217%) is transferred.