Question

Calculate the ratio of maximum wavelength of lyman and paschen series.

Answer 1

Answer:

7:108, huukhjekljrjrknjr

Answer 2

The ratio of maximum wavelength of lyman and paschen series is $\frac{7}{108}$

Explanation:

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

$\lambda$ = Wavelength of radiation

E= energy

For wavelength to be maximum, energy would be minimum.

Using Rydberg's Equation:

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$  = Higher energy level = 2

$n_i$= Lower energy level = 1 (Lyman series)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{lyman_{max}}}=R_H\left(\frac{1}{1^2}-\frac{1}{2^2} \right )$

$\lambda_{lyman_{max}}=\frac{4}{3R_H}$

2.

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$  = Higher energy level = 4

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

Putting the values, in above equation, we get

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

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

Thus $\frac{\lambda_{lyman}}{\lambda_{paschen}}=\frac{\frac{4}{3R_H}}{\frac{144}{7R_H}}=\frac{7}{108}$

Learn more about  Rydberg's Equation:

https://brainly.in/question/4783360

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