The ionisation potential of atomic hydrogen is 13.6 eV. A strong emission line in
the spectrum of atomic hydrogen occurs at a wavelength of 121.6 nm. Deduce the
initial and final values of the quantum number n for this spectral line.
To deduce the initial and final values of the quantum number n for this spectral line, we can use the Rydberg formula:
1/λ = R (1/n1^2 - 1/n2^2)
where λ is the wavelength of the emission line, R is the Rydberg constant (1.097373 x 10^7 m^-1), and n1 and n2 are the initial and final values of the quantum number n, respectively.
First, we need to convert the given wavelength from nanometers (nm) to meters (m):
121.6 nm = 121.6 x 10^-9 m
Substituting the values into the Rydberg formula:
1/(121.6 x 10^-9 m) = 1.097373 x 10^7 m^-1 (1/n1^2 - 1/n2^2)
Simplifying, we get:
1/(121.6 x 10^-9 m) = 1.097373 x 10^7 m^-1 (n2^2 - n1^2)
We can now rearrange the equation to solve for n2^2 - n1^2:
n2^2 - n1^2 = (1.097373 x 10^7 m^-1) / (121.6 x 10^-9 m)
n2^2 - n1^2 = 9
Now, we need to find two whole numbers whose difference is 9 and whose squares satisfy this equation. The only possible values for n2 and n1 are:
n2 = 4
n1 = 3
Therefore, the initial quantum number (n1) for this spectral line is 3, and the final quantum number (n2) is 4.