Examples for quantum physics
Example #1
Problem:
a.) What is the energy of a single photon (in eV) from a light source with a wavelength
of 400 nm?
Solution:
Use E = pc = hc/This gives anser in Joules, then multiply by 1.6E-19 to get answer in
eV.
3.1 eV
b.) If a 50 W laser emits 400 nm light, how many photons are emitted in 10 seconds?
Solution:
In 10 seconds, 500 Joules=500/1.6E-19 eV of photons are emitted. Dividing this total
energy by the energy per photon gives the total number of photons.
1.01E21
Example #2
Problem:
a.) Suppose light of wavelength 400 nm is incident on a metal with a work function W =
5.5 V. What external voltage must be applied to the metal to get electrons to be
released from the surface?
Solution:
From the previous problem, the energy of a single 400 nm photon is 3.1 eV. One must
therefore reduce the effective work function to 3.1 eV to allow the light to liberate an
electron.
2.4 eV
Example #3
Problem:
a.) A completely ionized Carbon nucleus is accelerated through a potential difference of
7000 volts. What is the final kinetic energy of the carbon? DATA: The charge of carbon
is 6 e and the mass is 12 proton masses.
Solution:
Use KE = qV. The charge is 6 e, so the answer in eV is
KE = 42.0 keV
b.) What is the DeBroglie wavelength of the Carbon?
Solution:
Use p = h/. But first one must use KE = p2/2m to find the momentum. (Remember to
get energy in Joules first)
p = sqrt(2*m*E) =1.64E-20 kg*m/sec.
 = 4.04E-14 m
Example #4
Problem:
a.) A monoenergetic beam of marbles which have a mass of 5.0 g is hurled into a board
with two slits. The velocity of the marbles is 15.0 m/sec, and the slits are separated by
6.0 cm. How far from the slits must one place a screen to get an interference pattern
where the first interference maxium is 20 cm from the central peak?
Solution:
First find the wavelength using p = mv = h/. One obtains a very short wavelength, =
8.84E-33 m. Since the angle will be small, one can use the relation y/L = /d where y is
20 cm and d is 6.0 cm. Solving for the length to the screen L,
L = 1.36E30 m = 1.4E14 light years
Example #5
Problem:
a.) An electron is confined to a box of length 0.6 nm (a typical atomic size). From the
uncertainty principle, estimate the minimum kinetic energy (in eV) of the electron.
Solution:
The momentum must be of order h/L. One can then estimate the kinetic energy with KE
= p2/2m. Remember to change the answer to eV.
KE = 8.4 eV