Question

Last line of bracket series for hydrogen atom has wavelength lambda1 and 2nd line of Lyman series has wavelength lambda 2 then find relation between lambda1 and lambda2?

Answer 1

Answer: Ratio of $\frac{\lambda_{1}}{\lambda_{2}}$ is 128:9

Explanation:  

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

$\lambda$ = Wavelength of radiation

E= energy

For the last line of Lyman series,  the electron will jump from n=1 to infinite level.

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 = $\infty$   (last line)

$n_i$= Lower energy level = 4 (Bracket series)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{lyman}}=R_H\left(\frac{1}{4^2}-\frac{1}{\infty^2} \right )$

$\lambda_{1}=\frac{16}{R_H}$

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 = 3 (second line)

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

Putting the values, in above equation, we get

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

$\lambda_{2}=\frac{9}{8\times R_H}$

Thus $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\frac{16}{R_H}}{\frac{9}{8\times R_H}}=\frac{128}{9}$