Question

For a gas of neutral hydrogen atoms, at what temperature will equal numbers of atoms have electrons in the a) ground state (n = 1) and in the first excited state (n = 2) and b) the ground state (n = 1) and second excited state (n = 3)?

Show all work.

Answer 1

Calculate number of energy state associated with n=1

g1= 2n^2 = 2 x 1 = 2

For n=2

g2= 2*(2)^2=8

We use the condition given in the problem and set up the equation

P(E_2)/P(E_1) =1 = (g_2/g_1) x exp -( E_2-E_1) /kT

Here P is the fraction k is the Boltzmann constant.

Lets calculate the value of E_2 and E_1

E_2 = -13.6 x ( 1/2^2)=. -(13.6/4)=-3.4 eV

E_1=-13.6 x ( 1/1)=-13.6 eV

Lets plug in these values to get the value of T.

1 = (8/2) x exp(-(-3.4-(-13.6)/8.6173 x 10^-5 eV x T)

(k = 8.6173 x 10^-5 eV/K )

Les solve for T

Exp ( -118366.5 /T )= 0.25

Lets take antilog of both side

( -118366.5 /T )= ln 0.25

( -118366.5 /T )= -1.38629

T = 85383.4 K

So the T = 85383.4 K

Part

Lets calculate for n=1 and n=3

E_3= -13.6 x ( 1/9)

-13.6/9=-1.511 eV

E_1=-13.6 eV

g_3= 2 x 9=18

g_1=2

Lets find T

1 = (18/2) x exp(-(-1.511-(-13.6)/8.6173 x 10^-5 eV x T)

0.1111 = exp(-(-1.511-(-13.6)/8.6173 x 10^-5 eV x T)

Lets take antilog of both side

Ln ( 0.111) = -(-1.511-(-13.6)/8.6173 x 10^-5 eV x T)

-2.19722 = -12.089/8.6173E-5 T

T = 140278.6/2.19722=63843.7 K