Physics, asked by MaryJane7512, 1 year ago

What will be the ratio of the wavelength of the first line to that of second line of paschen series?

Answers

Answered by QGP
193

In Hydrogen atom, the wavelength of radiation when an electron jumps between energy levels m and n is given by:




\displaystyle \frac{1}{\lambda} = R \left( \frac{1}{m^2}-\frac{1}{n^2}\right)




The different spectral series in Hydrogen Atom, in sequence are:



\boxed{\boxed{\begin{minipage}{11.5em}\begin{tabular}{c|c}\textbf{Series}&\textbf{Value of m}\\\cline{1-2}Lyman&1\\Balmer&2\\Paschen&3\\Brackett&4\\Pfund&5\end{tabular}\end{minipage}}}



For Paschen Series, the formula for wavelength becomes:



\displaystyle \frac{1}{\lambda} = R \left( \frac{1}{3^2}-\frac{1}{n^2}\right)




The value of n can be now 4,5,6,...



We have to find the ratio of wavelength of first line to that of second line of Paschen Series.



First line of Paschen Series is obtained by n=4.



\displaystyle \frac{1}{\lambda_1} = R \left( \frac{1}{3^2}-\frac{1}{4^2}\right)\\\\\\\implies \frac{1}{\lambda_1}=R\left(\frac{1}{9}-\frac{1}{16}\right)\\\\\\\implies \frac{1}{\lambda_1}= R \times \frac{7}{144}\\\\\\\implies \lambda_1=\frac{144}{7R} \quad ---(1)



The Second line of Paschen Series is obtained by n=5.



\displaystyle \frac{1}{\lambda_2} = R \left( \frac{1}{3^2}-\frac{1}{5^2}\right)\\\\\\\implies \frac{1}{\lambda_2}=R\left(\frac{1}{9}-\frac{1}{25}\right)\\\\\\\implies \frac{1}{\lambda_2}= R \times \frac{16}{225}\\\\\\\implies \lambda_2=\frac{225}{16R} \quad ---(2)




The ratio can be obtained by Dividing (1) by (2)




\displaystyle \frac{\lambda_1}{\lambda_2}=\frac{\frac{144}{7R}}{\frac{225}{16R}}\\\\\\\implies \frac{\lambda_1}{\lambda_2} = \frac{144}{7} \times \frac{16}{225}\\\\\\\implies\frac{\lambda_1}{\lambda_2}=\frac{16\times\cancel{9}}{7}\times\frac{16}{25\times\cancel{9}}\\\\\\\implies \boxed{\boxed{\bold{\frac{\lambda_1}{\lambda_2}=\frac{256}{175}}}}




Thus, The ratio of wavelength of first line to that of second line of Paschen Series is \bold{\frac{256}{175}}



tiwaavi: Great answer Indeed !!
QGP: Thank You Sir :)
HarishAS: Wow, Perfect answer bro.
QGP: Thank You Harish :)
aisha6067: Thanks for the correct answer
Answered by gunjansaini5746
16

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