A. The wavelength of the yellow light emitted by the 3p-3s transition in sodium

Question

A. The wavelength of the yellow light emitted by the 3p-3s transition in sodium is 590 nm. What is the energy of the emitted photon?

2.1000 eV

B. The probability of finding an atom at an excited state (relative to the probability of finding the atom in the ground state) is given by P=e(EkBT) At room temperature (T 300 K), what is the relative probability (compared to being in the ground state) that a sodium atom has its electron excited up to the 3p state?

P3p/P3s =

C. At the temperature of the flame (T 1500 K), what is the probability that a sodium atom has its electron excited up to the 3p state?

P3p/P3s = 6.9*10^-13

D. If we have 1 gram of sodium in our spoon, about how many atoms in the burner flame have their electron excited up to the 3p state at any one time? (Possibly useful fact: the atomic weight of sodium is 23 Da = 23 AMU.)

1.5*10^10

Explanation / Answer

(b)relative probablity= 1/(eE/KbT+1)

p= 1/(e2.1(1.6x10^-19)/1.38x10^-23 *300+1)

p=5.66x10-36 .

(C) relative probablity= 1/(eE/KbT+1)

p= 1/(e2.1(1.6x10^-19)/1.38x10^-23 *1500+1)

p=8.92x10-8 .

(d) Number of atoms in the excited state= (Na/23) * P

at 1500 k, N= (6.022x1023 /23)*8.92x10-8

N=2.34x1015

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