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Expert-verified Found in: Page 1218 ### Fundamentals Of Physics

Book edition 10th Edition
Author(s) David Halliday
Pages 1328 pages
ISBN 9781118230718 # (a) What is the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines? (b) What is the wavelength of the series limit?

(a) The wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines is 658 nm.

(b) Thus, the wavelength of the series limit is 366 nm.

See the step by step solution

## Step 1: Identification of the given data

The given data is listed below as-

The photon emitted in the Balmer series is the least energetic.

## Step 2: The energy equation is given by

The energy difference is given by the equation-

${\mathbf{∆}}{\mathbit{E}}{\mathbf{=}}\left(13.6eV\right)\left(\frac{1}{{n}_{2}^{2}}-\frac{1}{{n}_{1}^{2}}\right)$

Here, n is the quantum number.

## Step 3: To determine the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines (a)

The difference between energies is given by the equation:

$∆E={E}_{3}-{E}_{2}\phantom{\rule{0ex}{0ex}}∆E=-\left(13.6eV\right)\left(\frac{1}{{n}_{2}^{2}}-\frac{1}{{n}_{1}^{2}}\right)$

For, ${n}^{2}=3$ and ${n}_{1}=2$

$∆E=-\left(13.6eV\right)\left(\frac{1}{{3}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=1.889eV$

Now, hc = 1240 eV.nm

$\lambda =\frac{hc}{∆E}\phantom{\rule{0ex}{0ex}}=\frac{1240eV.nm}{1.889eV}\phantom{\rule{0ex}{0ex}}=658nm$

Thus, the wavelength of light for the least energetic photon emitted in the Balmer series of the hydrogen atom spectrum lines is 658 nm.

## Step 4: To determine the wavelength of the series limit. (b)

The difference between energies is given by the equation:

$∆E={E}_{\infty }-{E}_{2}\phantom{\rule{0ex}{0ex}}∆E=-\left(13.6eV\right)\left(\frac{1}{{n}_{2}^{2}}-\frac{1}{{n}_{1}^{2}}\right)$

For, ${n}_{2}=\infty$ and ${n}_{1}=2$

$∆E=-\left(13.6eV\right)\left(\frac{1}{{\infty }^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=3.40eV$

Now, hc = 1240 eV.nm

$\lambda =\frac{hc}{∆E}\phantom{\rule{0ex}{0ex}}=\frac{1240eV.nm}{3.40eV}\phantom{\rule{0ex}{0ex}}=366nm$

Thus, the wavelength of the series limit is 366 nm. ### Want to see more solutions like these? 