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

The wavelength of electron transition from n = 2 to n = 1 for Li2+ ion is obtained by dividing the wavelength for hydrogen by 9 because of the squared relationship to the atomic number in the Rydberg formula.

Step-by-step explanation:

The question deals with the comparison of electron transition wavelengths in a hydrogen atom and a lithium ion (Li2+). The Rydberg equation for hydrogen-like atoms, which include Li2+ due to having only one electron, expresses the wavelength (λ) of emitted or absorbed light in terms of the atomic number (Z) and the principal quantum numbers (ni and nf) as 1/λ = RZ2 (1/nf2 - 1/ni2), where R is the Rydberg constant. Since the atomic number of hydrogen is 1 and that of lithium is 3, and the transition in both cases is from n = 2 to n = 1, we can say that for lithium ion, the wavelength will be affected by the square of the atomic number. Therefore, you can find the wavelength for Li2+ by dividing the wavelength for hydrogen by 32 or 9.

The correct answer to the question "By what coefficient should this wavelength be multiplied to obtain the wavelength associated with the same electron transition in the Li2+ ion?" is E) 1/9.

Answer 2

The correct answer is option E.

Step-by-step explanation:

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

Using Rydberg's Equation:

`(1)/(\lambda)=R_H* Z^2\left((1)/(n_f^2)-(1)/(n_i^2) \right )`

Where,

`\lambda` = Wavelength of radiation

`R_H=1.097* 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* 10^(-7) m`

Putting the values, in above equation, we get

`(1)/(\lambda)=R_H* 1^2\left((1)/((1)^2)-(1)/((2)^2) \right )`

`(1)/(\lambda)=R_H* (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

`(1)/(\lambda ')=R_H* 3^2\left((1)/((1)^2)-(1)/((2)^2) \right )`

`(1)/(\lambda ')=R_H* 9* (3)/(4)`..[2]

Dividing [1] and [2]

`((1)/(\lambda ))/((1)/(\lambda '))=(R_H* (3)/(4))/(R_H* 9* (3)/(4))`

`\lambda '=(1)/(9)* \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.