Question

The wavelength of light associated with the n = 2 to n = 1 electron transition in the hydrogen spectrum is 1.216 × 10^(–7) m.
By what coefficient should this wavelength be multiplied to obtain the wavelength associated with the same electron transition in the Li^(2+) ion?

A) 1/3 B) 1/4 C) 1/7 D) 1 E) 1/9

Answer 1

Answer:

The correct answer is option E.

Explanation:

For wavelength to be minimum, energy would be maximum, i.e the electron  will jump to infinite level.

Using Rydberg's Equation:

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

Where,

$\lambda$ = Wavelength of radiation

$R_H=1.097\times 10^7 m^{-1}$ = Rydberg's Constant

$n_f$ = Higher energy level

$n_i$= Lower energy level

Z = atomic number

For hydrogen , Z = 1

The wavelength of light associated with the n = 2 to n = 1 electron transition in the hydrogen spectrum= $\lambda =1.216\times 10^{-7} m$

Putting the values, in above equation, we get

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

$\frac{1}{\lambda}=R_H\times \frac{3}{4}$..[1]

For $Li^{2+}$,

Z = 3

The wavelength of light associated with the n = 2 to n = 1 electron transition in the lithium ion = $\lambda '=?$

Putting the values, in above equation, we get

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

$\frac{1}{\lambda '}=R_H\times 9\times \frac{3}{4}$..[2]

Dividing [1] and [2]

$\frac{\frac{1}{\lambda }}{\frac{1}{\lambda '}}=\frac{R_H\times \frac{3}{4}}{R_H\times 9\times \frac{3}{4}}$

$\lambda '=\frac{1}{9}\times \lambda$

The coefficient required to be multiplied with wavelength of transition given for hydrogen spectrum  to obtain the wavelength associated with the same electron transition in the $Li^{2+}$ ion is 1/9.