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soldi70 [24.7K]
3 years ago
6

Show that the entire Paschen series is in the infrared part of the spectrum. To do this, you only need to calculate the shortest

wavelength in the series.
Physics
1 answer:
mr_godi [17]3 years ago
8 0

Answer and Explanation:

The computation of the shortest wavelength in the series is shown below:-

\frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )

Where

\lambda represents wavelength

R represents Rydberg's constant

n_f represents Final energy states

and n_i represents initial energy states

Now Substitute is

1.097\times 10^7\ m^{-1}\ for\ R, \infty for\ n_i,\ 3 for\ n_i,\\\\\ \frac{1}{\lambda} = R(\frac{1}{n_f^2} - \frac{1}{n_i^2} )

now we will put the values into the above formula

= 1.097\times 10^7 m^{-1}(\frac{1}{3^2} - \frac{1}{\infty^2} )\\\\ = 1.097\times10^7\ m^{-1} (\frac{1}{9} )

= 1218888.889 m^{-1}

Now we will rewrite the answer in the term of \lambda

\lambda = \frac{1}{1218888.889} m\\\\ = 0.82\times 10^{-6} m

So, the whole Paschen series is in the part of the spectrum.

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  • \theta = 20^\circ (or, equivalently, (\pi / 9) radians, if the calculator is in radian mode.)
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