Question

Calculate the shortest and longest wavelength of balmer series of hydrogen atom.

Answer 1

Wavelength and energy are inversely proportional so as the energy increase the wavelength decrease and vice versa. All the balmer series of hydrogen atom has n=2

Calculating the wavelenght using rydberg equation and the rydberg's equation is:

For longer wavelength:

$\frac{1}{lambda}$ =$R =(\frac{1}{n_{f}^{2}  } -\frac{1}{n_{i}^{2}  } )$

= $(1.907*10^{7}m^{-1})(\frac{1}{2^{2} } -\frac{1}{3^{2} } )$

= 1.52$*10^{6}$$m^{-1}$

Therefore $\frac{1}{lambda} = 1.52*10^{6}m$

lambda =  $\frac{1}{1.52*10^{6} }m$

             = 6.56*$10^{-7}$m = 656nm

For shorter wavelength:

$\frac{1}{lambda} = R(\frac{1}{n_{f}^{2}  } -\frac{1}{n_{i}^{2}  } )$

= $\frac{1}{lambda} = (1.907*10^{7})(\frac{1}{2^{2} } -\frac{1}{∞^{2} } )$

therefore lambda = $3.65*10^{-7} m$ =365nm

Answer 2

Answer:

Wavelength and energy are inversely proportional so as the energy increase the wavelength decrease and vice versa. All the balmer series of hydrogen atom has n=2

Calculating the wavelenght using rydberg equation and the rydberg's equation is:

For longer wavelength:

 =

= 

= 1.52

Therefore 

lambda =  

            = 6.56*m = 656nm

For shorter wavelength:

= 

therefore lambda =  =365nm

Explanation: