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