Question

Calculate the smallest wavelength of radiation that may be emitted by (a) hydrogen, (b) He⁺ and (c) Li⁺⁺.

Answer 1

Given:

For the smallest wavelength, energy should be maximum.

Thus, for maximum energy, transition should be from infinity to the ground state.

`therefore n_1 = 1`

`n_2 = ∞`

(a) Wavelength of the radiation emitted`(lamda)` is given by

`1/lamda = RZ^2 (1/n_1^2 - 1/n_2^2 )`

For hydrogen,

Atomic number, Z = 1

R = Rydberg constant = `1.097xx10^7m^-1`

On substituting the respective values, 

   
`lamda = 1/(1.097xx10) = 1/1.097xx 10^-7`

= 0.911 × 10^-7

= 91.01 × 10^-9 = 91 nm

(b)

For He⁺,

Atomic number, Z = 2

Wavelength of the radiation emitted by He⁺ `(lamda)` is given by

`1/lamda = RZ^2 (1/n_1^2 - 1/n_2^2)`

`therefore 1/lamda = (2)^2 (1.097xx10^7 )(1/((1^2)) - 1/((∞^2))` 

`therefore lamda = (91 nm)/4 =23  nm`

(c) For Li^++ ,

 Atomic number, Z = 3
Wavelength of the radiation emitted by  Li⁺⁺  (λ) is given by

`1/lamda = RZ^2 (1/n_1^2 - 1/n_2^2)`

`therefore 1/lamda = (3)^2 xx (1.097 xx 10^7)   (1/1^2 - 1/∞^2)`

`rArr lamda = (91 nm)/Z^2 = 91/9 = 10 nm `