Question

Q] Calculate the wavelength of first three lines in paschen series of hydrogen atom​

Answer 1

Question :

>> Calculate the wavenumber of the first line in paschen series of hydrogen spectrum.

Solution:

wave number = R( 1/n1^2 - 1/n2^2)

= 109677 ( 1 / 3^2 - 1/4^2)

= 109677 ( 1/9 - 1/16)

= 109677 ( 16 - 9 / 144 )

= 761.64 × 7

= 5331.48

hope it helps you

Answer 2

Answer:

☆Wavelength of paschen series:

》when spectral lines are falling from infinity to 3rd energy level and it goes on...lasted from 4th energy level to 3rd.

》so the value of 'n' for first three lines are...4,5,6.

☆Wavelength calculation:

• $\boxed{\bf{\dfrac{1}{\lambda} = R_{H}(\dfrac{1}{3^{2}}- \dfrac{1}{n^{2}})}}$

where $\ R_{H}$ is rydberg's constant = $\ 1.1×10^{7}$

☆solution:

》Wavelength of first line, n=4

$\rightarrow{\sf{\dfrac{1}{\lambda_{1}} = R_{H}(\dfrac{1}{3^{2}}- \dfrac{1}{n^{2}})}}$

$\rightarrow{\sf{\dfrac{1}{\lambda_{1}} = 1.1×10^{7}(\dfrac{1}{3^{2}}- \dfrac{1}{4^{2}})}}$

$\rightarrow{\sf{\dfrac{1}{\lambda_{1}} = 1.1×10^{7}(\dfrac{1}{9}- \dfrac{1}{16})}}$

$\rightarrow{\sf{\dfrac{1}{\lambda_{1}} = 1.1×10^{7}(\dfrac{7}{144})}}$

$\rightarrow{\sf{\lambda_{1} = \dfrac{144×10^{-7}}{7.7}}}$

$\rightarrow{\bf{\lambda_{1} = 1870nm}}$

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》Wavelength of 2nd line,n=5

$\rightarrow{\sf{\dfrac{1}{\lambda_{2}} = R_{H}(\dfrac{1}{3^{2}}- \dfrac{1}{5^{2}})}}$

$\rightarrow{\sf{\dfrac{1}{\lambda_{2}} = R_{H}(\dfrac{1}{9}- \dfrac{1}{25})}}$

$\rightarrow{\sf{\dfrac{1}{\lambda_{2}} = 1.1×10^{7}(\dfrac{16}{225})}}$

$\rightarrow{\sf{ \lambda_{2} =\dfrac{225×10^{-7}}{17.6}}}$

$\rightarrow{\bf{\lambda_{2} = 1258nm}}$

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》Wavelength of 3rd line,n=6

$\rightarrow{\sf{\dfrac{1}{\lambda_{3}} = R_{H}(\dfrac{1}{3^{2}}- \dfrac{1}{6^{2}})}}$

$\rightarrow{\sf{\dfrac{1}{\lambda_{3}} = R_{H}(\dfrac{1}{9}- \dfrac{1}{36})}}$

$\rightarrow{\sf{\dfrac{1}{\lambda_{3}} = 1.1×10^{7}(\dfrac{1}{12})}}$

$\rightarrow{\sf{\lambda_{3} = 1.090×10^{-6}}}$

$\rightarrow{\bf{\lambda_{3} = 1090nm}}$

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》so first 3 wavelength of paschen series are:

1. 1870nm
2. 1258nm
3. 1090nm

》conclusion: energy realeased increases during the transmission from 4...5..6..wavelength decreases.

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♧hope it helps!♧