Question

Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.

Answer 1

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Answer 2

Answer: The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen is $3\times 10^5m^{-1}$

Explanation:

$E=\frac{hc}{\lambda}$

$\lambda$ = Wavelength of radiation

E= energy

For wavelength to be maximum, energy would be minimum, i.e the electron will jump from n=2 to n=3  for balmer series.

Using Rydberg's Equation:

$\frac{1}{\lambda}=R_H\left(\frac{1}{n_i^2}-\frac{1}{n_f^2} \right )\times Z^2$

Where,

$\lambda$ = Wavelength of radiation  = ?

$R_H$ = Rydberg's Constant  =$10973731.6m^{-1}$

$n_f$ = Higher energy level = 3

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

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

$\frac{1}{\lambda}=10973731.6m^{-1}\left(\frac{1}{2^2}-\frac{1}{3^2} \right )\times 1^2$

$\frac{1}{\lambda}=3\times 10^5m^{-1}$

$\nu=\frac{1}{\lambda}$

Where,

$\lambda$ = Wavelength of radiation

$\nu$ = wavenumber

Thus $\nu=3\times 10^5m^{-1}$

The wave number for the longest wavelength transition in the Balmer series of atomic hydrogen is $3\times 10^5m^{-1}$