Question

Difference in wavelength of two extreme lines of Lyman series in emission spectrum of he + is

Answer 1

Answer: $\frac{1}{12R_H}$

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 =1  level to infinite level for Lyman series.

For wavelength to be be maximum, energy would be maximum, i.e the electron will jump from n=1 to n=2  for Lyman 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

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

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

Z= atomic number = 2 (for helium)

Putting the values, in above equation, we get

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

$\lambda_{lyman_{max}}=\frac{1}{4R_H}$

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

Z= atomic number = 2 (for helium)

Putting the values, in above equation, we get

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

$\lambda_{lyman_{min}}=\frac{1}{3R_H}$

Thus ${\lambda_{max}}-{\lambda_{min}}={\frac{1}{3R_H}}-{\frac{1}{4R_H}}=\frac{1}{12R_H}$

Thus difference in wavelength of two extreme lines of Lyman series in emission spectrum of $He^+$  is $\frac{1}{12R_H}$