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

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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−(ΔEkBT) 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?

P_3p/P_3s =

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?

P_3p/P_3s = 6.9*10^-13

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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

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Answer 1

(b)relative probablity= 1/(e^E/KbT+1)

p= 1/(e^{2.1(1.6x10^-19)/1.38x10^-23 *300}+1)

p=5.66x10^-36 .

(C) relative probablity= 1/(e^E/KbT+1)

p= 1/(e^{2.1(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