(a) Draw the energy level diagram for the line spectra representing Lyman series and Balmer series in the spectrum of hydrogen atom.

(b) Using the Rydberg formula for the spectrum of hydrogen atom, calculate the largest and shortest wavelengths of the emission lines of the Balmer series in the spectrum of hydrogen atom. (Use the value of Rydberg constant R=1.1\times 10^{7}m^{-1})

 

 

 
 
 
 
 

Answers (1)
S safeer

The energy level diagram for line spectra 

(a)

 

(b)

\\\frac{1}{\lambda _{max}}=R(\frac{1}{2^2}-\frac{1}{3^2})\\\\\frac{1}{\lambda _{max}}=R\frac{5}{36}

\\\lambda_{max}=\frac{36}{5\times 1.1\times 10^{7}}m        \left ( \because R=1.1\times 10^{7} \right )

\lambda _{max}=6.5\times 10^{-7}\ m is the largest wavelength of emission 

Now calculation for smallest wavelength, we have :

\frac{1}{\lambda _{smallest}}=R\left \{ \frac{1}{2^{2}}- \frac{1}{\infty ^{2}}\right \}

\frac{1}{\lambda _{smallest}}=\frac{R}{4}

\lambda _{min.}=3.6\times 10^{-7}m is the smallest wavelength of emission.

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