Question

. A kind of atom at an excited state has a life time of 7.8 ns. It relax to ground state by emitting one photon.

a. What is the probability that a such excited atom remains in the excited state after 2.0 ns?

b. What is the probability of a such excited atom will have emitted a photon within the first 10 ns interval?

c. If you start with 105 of such excited atoms, how many atoms will remain excited after 2.0 ns?

d. How many photons will have been emitted within the first 10 ns?

Answer 1

a] Use the Boltzmann population density distribution function

for a single atom (normalized system),

so, at t = 2ns,

this is the probability that the atom is in excited state at 2ns.

b] at t = 10ns,

so, probability that a photon is emitted by t=10ns is: 100 - 27.74 = 72.26%.

c] After 2ns, the number of atoms remaining in the excited state will be:

N = 105 x 0.7738 = 81.249 or approximately 81.

d] Number of photons emitted = number of atoms exited the excited state = 105 x 0.7226 = 75.873 or approximately 76.