Solutions for Recitation Exercise # 11
1. The term “quantum of light” means that a photon is the smallest ‘unit’ of light, carrying a definite
amount of energy. The photon’s energy is quantized.
2. Increasing the intensity of a beam of light increases the number of photons per second. (Answer iv)
3. Increasing the frequency of a beam of light increases the energy of each photon, without changing the
intensity. (Answer ii)
4. Since the electron and proton have the same (magnitude) charge, they will both see the same difference
in potential energy across a potential difference of 100 V (although downhill for one will be uphill for
the other). So starting from rest, accelerating them through this potential difference they will end up
with the same kinetic energy. (Yes, as some of you asked in class, they would accelerate through this in
opposite directions.)
√
√
Since K = p2 /2m, this means p = 2mK, so λ = h/ 2mK. The kinetic energies are the same, so the
electron has the larger de Broglie wavelength, since its mass is much smaller than the mass of the proton.
5. Since temperature is directly proportional to kinetic
√ energy, particles in beam 1 have a higher average
kinetic energy than in beam 2. Again using λ = h/ 2mK, this means that beam 2 has the larger de
Broglie wavelength.
6. The neutron will slow down as it ascends, so its kinetic energy will decrease. This means that its de
Broglie wavelength will increase as the neutron rises.
7. Since En = − 13.6n2eV , the energy level diagram will have E1 at −13.6 eV , with progressively higher
En ’s getting closer together as they approach E = 0. See picture in your textbook. For an electron in
the n = 4 state to emit a photon, it must transition to a lower energy level. (Energy is conserved, so
if the photon is created with some energy, the electron must have lost energy.) The possible downward
transitions here are n = 4 → 3, n = 4 → 2, n = 4 → 1.
8.a Here we’re assuming the Hydrogen atom is in the ground state before the collision (a pretty good
assumption). There are two electrons involved – the electron bound to the proton to form the atom, and
the electron that is coming in with kinetic energy of 12.5 eV and colliding with the atom. In order to
excite the atom, it needs to gain energy exactly corresponding to the energy difference from n = 1 to
n = 2, or n = 1 to n = 3, and so on. The most energy it could get would be 12.5 eV . It turns out that
this is enough to excite the Hydrogen atom to n = 2 or 3, but not 4 or higher. This is easiest to see if
you put the energies and not just the n’s in your diagram in the last problem.
[The significance of the H atom’s energy being negative is that this is a bound state of an electron and
proton. In PHY 121 you may remember that for the system of the Moon orbiting the Earth, for example,
the potential energy is negative. Negative PE means the same thing here. It is the potential energy of the
bound electron–proton system, and once the energy gets up to zero, the electron is free. For gravitational
systems, you used this to find the escape velocity – i.e. the speed which makes KE + P E = 0. Here, the
energy you need to add in order to free the electron is called the ionization energy.]
b After the collision, the electron will have given some of its energy to the atom, so it will (iii) bounce
off with K < 12.5 eV .
c After the collision, the atom could be in the n = 2 or the n = 3 state, so possible downward transitions
(emit a photon → lose energy → transition to a lower energy (lower n) state) are 3 → 2, 3 → 1, 2 → 1.
9.a The wavelength 656 nm corresponds to the 3 → 2 transition. You can do this using E = hc/λ for
the photon and En = −13.6/n2 eV for the atom, but it is easier if you use the given information that
this is the longest wavelength in the Balmer series. The largest wavelength means the smallest energy
photon. The “Balmer series” means “downward transitions which end on n = 2” and the nearest energy
level is n = 3. So this is the 3 → 2 transition.
b The Hydrogen atoms will be in the ground state. (Excited atoms eventually decay, and there is nothing
to keep these atoms excited. Thermal energy won’t be enough unless the temperature is really high, since
1
eV .) However, since this photon’s energy is exactly the difference between
at room temperature kT ≈ 40
the n = 2 and 3 states, the only way it would be absorbed is if it hits an atom already in n = 2. So the
answer is no, it will not be absorbed.