Question

The wavelength of the first line of lyman series for hydrogen atom is equal to that of the second line of balmer series for a hydrogen like ion. the atomic number z of hydrogen like ion is:

Answer 1

Here is the answer-

Check it below....

Answer 2

Answer: Atomic number Z of hydrogen like ion is $\frac{1}{2}$

Explanation:

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

$\lambda$ = Wavelength of radiation

E= energy

1. For first line in Lyman series, the electron will jump from n=1  level to n=2.

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = 2  

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

Z= atomic number = 1 (for hydrogen)

Putting the values, in above equation, we get

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

$\lambda_{lyman}=\frac{4}{3R_H}$

2. For second line in Balmer series, the electron will jump from n=2  level to n=4

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level = 4

$n_i$= Lower energy level = 2

Z= atomic number = ? (for hydrogen like ion)

Putting the values, in above equation, we get

$\frac{1}{\lambda_{balmer}}=R_H\left(\frac{1}{2^2}-\frac{1}{4^2} \right )\times Z^2$

$\lambda_{balmer}=\frac{16}{3R_H}\times Z^2$

Given : $\lambda_{lyman}=\lambda_{balmer}$

Thus $\frac{4}{3R_H}=\frac{16}{3R_H}\times Z^2$

$Z=\frac{1}{2}$

Thus the atomic number Z of hydrogen like ion is $\frac{1}{2}$