A moving helium atom absorbs a photon of wavelength 12.2 angstrom and stops. What was the speed of helium atom?
Answers
Well! We know that the energy of a quantum of photon is given by E=hf. We also know from waves that f=c/w where w is the wavelength. If we substitute this into the previous equation, we get E=hc/w.
If the helium atom stops after absorbing the photon, this means the the energy of the photon was exactly equal to the kinetic energy of the helium (so it cancels out). Therefore:
KE=hc/w
I’m going to assume that the helium atom is moving at non-relativistic velocities and that the kinetic energy of the atom is the classical kinetic energy (KE=1/2mv^2). We can plug this into the previous equation to get the new equation.
1/2mv^2=hc/w
Now we rearrange to solve for the velocity “v”.
v=sqrt(2hc/mw)
Because I assumed that the velocity was non-relativistic, this means that I can use the invariant mass (or rest mass) of a helium atom and plug in the other values as well. For mass, I will use kg, “c” will be in m/s and “w” will be in meters.
v=sqrt(2(6.63*10^-34)(3*10^8)/(2.66*10^-26)(1.22*10^9))
After the math…
v=1.12*10^-4 m/s
If anyone noticed that I did things improperly, please let me know