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Q49P

Expert-verifiedFound in: Page 1248

Book edition
10th Edition

Author(s)
David Halliday

Pages
1328 pages

ISBN
9781118230718

**Assume that lasers are available whose wavelengths can be precisely “tuned” to anywhere in the visible range—that is, in the range ${\mathbf{450}}{\mathbf{}}{\mathbf{nm}}{\mathbf{<}}{\mathbf{\lambda}}{\mathbf{<}}{\mathbf{650}}{\mathbf{}}{\mathbf{nm}}{\mathbf{}}$. If every television channel occupies a bandwidth of 10 MHz, how many channels can be accommodated within this wavelength range?**

There can be $2.1\times {10}^{7}$ channels accommodated within this wavelength range.

a) Wavelength range of the visible region, $450\mathrm{nm}<\mathrm{\lambda}<650\mathrm{nm}$

b) Bandwidth of the television channel, $\Delta f=10\mathrm{MHz}$

**Using the formula of energy due to Planck's relation, we can get the energy difference value of the states. Now, for the corresponding energy difference between the two states, we can consider the energy difference between the states of the hydrogen atom n = 1 and n = 2 to compare the difference value.**

Formula:

The frequency of a wave,

$f=\frac{c}{\lambda}$ ….. (1)

Here, $\lambda $ is the wavelength.

Speed of light,$c=3\times {10}^{8}\frac{m}{s}$

Let, the range of frequency of the microwave be $\u2206f$.

Then, the number of channels that could be accommodated within the range can be given as follow.

$N=\frac{\u2206f}{10MeV}\phantom{\rule{0ex}{0ex}}=\frac{1}{{10}^{7}eV}\left(\frac{3\times {10}^{8}m/s}{450\times {10}^{-9}m}-\frac{3\times {10}^{8}m/s}{650\times {10}^{-9}m}\right)\phantom{\rule{0ex}{0ex}}=6.7\times {10}^{7}-4.6\times {10}^{7}\phantom{\rule{0ex}{0ex}}=2.1\times {10}^{7}\mathrm{channels}$

$\mathrm{N}=21.0\times {10}^{6}\mathrm{channels}\phantom{\rule{0ex}{0ex}}=21.0\mathrm{million}\mathrm{channels}$

Hence, the number channels accommodated is $2.1\times {10}^{7}$.

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