Question

Calculate the longest and the shortest wavelength observed in the Balmer series of the H atom spectrum

Answer 1

The longest wavelength observed in the Balmer series of the H atom spectrum is 656.3 nm.

The shortest wavelength observed in the Balmer series of the H atom spectrum is 364.6 nm.

Step-by-step explanation:

Using Rydberg's Equation:

`(1)/(\lambda)=R_H\left((1)/(n_i^2)-(1)/(n_f^2) \right )`

Where,

`\lambda` = Wavelength of radiation

`R_H` = Rydberg's Constant

`n_f` = Higher energy level

`n_i`= Lower energy level

`(1)/(\lambda)=R_H\left((1)/(n_i^2)-(1)/(n_f^2) \right )`

For wavelength to be longest, energy would be minimum, i.e the electron will jump from third level to second level :

`n_f` = Higher energy level =
`3`

`n_i`= Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

`(1)/(\lambda_(Balmer))=R_H\left((1)/(2^2)-(1)/(3^2) \right )`

`\lambda_(Balmer)=(36)/(5R_H)`

`\lambda_(Balmer)=(36)/(5* 1.097* 10^7 m^-1)}=6.563* 10^(-7) m=656.3 nm`

`1 m =10^9 nm`

The longest wavelength observed in the Balmer series of the H atom spectrum is 656.3 nm.

For wavelength to be shortest, energy would be maximum, i.e the electron will from infinite level to second level. :

`n_f` = Higher energy level =
`\infty`

`n_i`= Lower energy level = 2 (Balmer series)

Putting the values, in above equation, we get

`(1)/(\lambda_(Balmer))=R_H\left((1)/(2^2)-(1)/(\infty^2) \right )`

`\lambda_(Balmer)=(4)/(R_H)`

`=(4)/(1.097* 10^7 m^-1)=3.646* 10^(-7) m=364.6 nm`

The shortest wavelength observed in the Balmer series of the H atom spectrum is 364.6 nm.