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Q62P

Expert-verifiedFound in: Page 1218

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.

The given data is listed below as-

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

**The ****energy difference**** is given by the equation-**

${\mathbf{\u2206}}{\mathit{E}}{\mathbf{=}}{\left(13.6eV\right)}{\left(\frac{1}{{n}_{2}^{2}}-\frac{1}{{n}_{1}^{2}}\right)}$

Here, n is the quantum number.

The difference between energies is given by the equation:

$\u2206E={E}_{3}-{E}_{2}\phantom{\rule{0ex}{0ex}}\u2206E=-\left(13.6eV\right)\left(\frac{1}{{n}_{2}^{2}}-\frac{1}{{n}_{1}^{2}}\right)$

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

$\u2206E=-\left(13.6eV\right)\left(\frac{1}{{3}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=1.889eV$

Now, hc = 1240 eV.nm

$\lambda =\frac{hc}{\u2206E}\phantom{\rule{0ex}{0ex}}=\frac{1240eV.nm}{1.889eV}\phantom{\rule{0ex}{0ex}}=658nm$

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

The difference between energies is given by the equation:

$\u2206E={E}_{\infty}-{E}_{2}\phantom{\rule{0ex}{0ex}}\u2206E=-\left(13.6eV\right)\left(\frac{1}{{n}_{2}^{2}}-\frac{1}{{n}_{1}^{2}}\right)$

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

$\u2206E=-\left(13.6eV\right)\left(\frac{1}{{\infty}^{2}}-\frac{1}{{2}^{2}}\right)\phantom{\rule{0ex}{0ex}}=3.40eV$

Now, hc = 1240 eV.nm

$\lambda =\frac{hc}{\u2206E}\phantom{\rule{0ex}{0ex}}=\frac{1240eV.nm}{3.40eV}\phantom{\rule{0ex}{0ex}}=366nm$

Thus, the wavelength of the series limit is 366 nm.

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