Question

Calculate the maximum and minimum wavelength in the paschen series in hydrogen spectrum

Answer 1

Minimum wavelength : 8199A°

Maximum wavelength : 18741A°

Answer 2

The maximum and minimum wavelength in the paschen series in hydrogen spectrum are $\frac{144}{7R_H}$  and $\frac{9}{R_H}$  respectively

Explanation:

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

$\lambda$ = Wavelength of radiation

E= energy

For wavelength to be minimum, energy would be maximum, i.e the electron will jump from = 3  level to infinite level for Paschen series.

For wavelength to be be maximum, energy would be minimum, i.e the electron will jump from n=3 to n=4  for Paschen 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,

$\lambda$ = min Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = $\infty$  

$n_i$= Lower energy level = 3 (Paschen series)

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{paschen}}=R_H\left(\frac{1}{3^2}-\frac{1}{\infty^2} \right )\times 1$

$\lambda_{paschen_{min}}=\frac{9}{R_H}$

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

Where,

$\lambda$ = max Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = 4

$n_i$= Lower energy level = 3

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{paschen_{max}}}=R_H\left(\frac{1}{3^2}-\frac{1}{4^2} \right )\times 1^2$

$\lambda_{paschen_{max}}=\frac{144}{7R_H}$

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