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 Li2+ ion?
Go through each step, and explain in detail.

Answer 1

Answer:

The coefficient should be 0.1111.

Explanation:

Using Rydberg's Equation:

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

Where,

$\lambda$ = Wavelength of radiation

$R_H$ = Rydberg's Constant

$n_f$ = Higher energy level

$n_i$= Lower energy level

Z = Atomic number

1) For n = 2 to n = 1 in hydrogen atom:

Z = 1

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

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

2) For n = 2 to n = 1 in lithium ion i.e. $Li^{2+}$ :

Z = 3

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

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

[2] ÷ [1]

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

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

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

$\lambda '=\lambda \times 0.1111$

The coefficient should be 0.1111.

Answer 2

0.1111 is the coefficient that should the wavelength be multiplied to obtain the wavelength associated with the same electron transition in the Li2+ ion

What is the wavelength of light?

The wavelength of light means the distance between the crest and trough of light.

By Rydberg's equation

$\dfrac{1}{\lambda} = R_H \times Z^2 (\dfrac{1}{n^2i} -\dfrac{1}{n^2f} )$

For n = 2 to n = 1 in hydrogen atom

Z = 1

$\dfrac{1}{\lambda} = R \times 1^2 (\dfrac{1}{2^2} -\dfrac{1}{1^2} )\\\\\\\dfrac{1}{\lambda} = -R \times \dfrac{3}{4}$

For n = 2 to n = 1 in lithium-ion

Z = 3

$\dfrac{1}{\lambda} = R \times 3^2 (\dfrac{1}{2^2} -\dfrac{1}{1^2} )\\\\\\\dfrac{1}{\lambda'} = -9R \times \dfrac{3}{4} \\\\\\\dfrac{1/\lambda'}{1/\lambda} =\dfrac{ -9R \times3/4}{-R \times3/4} = 0.1111$

Thus, the coefficient is 0.1111

Learn more about wavelength, here:

https://brainly.com/question/7143261