Question

Figure 37.7 identified the wavelengths of four lines in the Balmer series of hydrogen.

a. Determine the Balmer formula n and m values for these wavelengths.

b. Predict the wavelength of the fifth line in the spectrum .

Short answer

(a) The Balmer formula is$\lambda =\frac{91.18}{{\displaystyle \frac{1}{m^2}}-{\displaystyle \frac{1}{n^2}}}\mathit{n}\mathit{m}$

(b) Wavelength of the fifth line in the spectrum =$397.139nm$

Step-by-step solution

Part(a) Step 1: Given information

We have given the emission spectrum of hydrogen for four different

wavelengths in fig 37.7 .

We have to find the Balmer formula with n and m values of given wavelengths.

Part(a) Step2: Simplify

Balmer showed that the wavelengths in the hydrogen spectrum could be represented by the simple formula,

where ;$m=1,2,3....$

$n=anyintegergreaterthanm$

$\lambda =\frac{91.18}{{\displaystyle \frac{1}{m^2}}-{\displaystyle \frac{1}{n^2}}}\mathit{n}\mathit{m}$

This formula predicts a series of spectral lines of gradually decreasing wavelength, converging to the series limit wavelength of 364.7 nm as n tends to ∞.

This series of spectral lines is now called the Balmer series.

Part(b) Step1: Given information

Balmer series of hydrogen spectrum.

We have to find the wavelength for the fifth line of Balmer series spectrum.

Part(b) Step2 : Simplify                   

From Balmer formula ,

$\lambda =\frac{91.18}{{\displaystyle \frac{1}{m^2}}-{\displaystyle \frac{1}{n^2}}}\mathit{n}\mathit{m}$

We know Balmer series starts from $m=2$

So ,the fifth line of the Balmer series spectrum is from $m=2ton=7$

By substituting the value of $m=2$and $n=7$in Balmer formula ,we get

$$\begin{gathered} \lambda =\frac{91.18}{{\displaystyle \frac{1}{2^2}}-{\displaystyle \frac{1}{7^2}}}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{1}{4}}-{\displaystyle \frac{1}{49}}}\mathit{n}\mathit{m} \\ =\frac{91.18}{{\displaystyle \frac{49-4}{49\times 4}}}\mathit{n}\mathit{m} \\ =397.139\mathbf{}\mathit{n}\mathit{m} \end{gathered}$$