Sol. We can use the equation

= R with n l = 1, n u = 2

= R = λ = =

The frequency of the photon is f = = =

2.47 × 10 15 Hz The energy of the photon is

E = hf = (4.136 × 10 –15 )(2.47 × 10 15 ) = 10.2 eV

From conservation of momentum, as the total

momentum before emission is zero, the total

momentum after emission must also be zero. The

photon and atom therefore move off in opposite

directions, with

mv =

where m and v are the mass and recoil speed of the hydrogen atom, E photon is the actual energy of the photon (less than 10.2 eV) and c is the speed of light. The energy difference between the n = 2 and n = 1 levels, E is the source of both the photon energy and the recoil kinetic energy of the atom. From energy

conservation, we have

E = E photon + mv 2 atom being massive, we can assume that its recoil speed v and kinetic energy are so small that E = E photon . Substituting E photon = 10.2 eV into the expression for mv yields mv = 10.2 eV/c

The recoil kinetic energy of the hydrogen atom can now be calculated.

K = mv

2 = = (0.5) = =

5.54 × 10 –8 eV

Thus the fraction of the energy difference between the

n = 2 and n = 1 levels that goes into atomic recoil

energy is very small. = = 5.43 × 10 –9 That is why by equating the photon’s energy to the

atomic energy level separation yields accurate answers because little energy is needed to conserve momentum.