Examples for quantum physics

Example #1


a.) What is the energy of a single photon (in eV) from a light source with a wavelength of 400 nm?


Use E = pc = hc/l. 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?


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.


Example #2


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?


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


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.


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?


Use p = h/l. 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.

l = 4.04E-14 m

Example #4


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?


First find the wavelength using p = mv = h/l. One obtains a very short wavelength, l = 8.84E-33 m. Since the angle will be small, one can use the relation y/L = 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


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.


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

Quantum physics' index