An electron of wavelength 1.74*10-10m strikes an atom of ionized helium (He+). What is the wavelength (m) of the light corresponding to the line in the emission spectrum with the smallest energy transition?
To determine the wavelength of the light corresponding to the line in the emission spectrum with the smallest energy transition in helium ion (He+), we need to use the Rydberg formula.
The Rydberg formula is given by:
1/λ = R * (1/n1^2 - 1/n2^2)
Where:
- λ is the wavelength of the light
- R is the Rydberg constant (approximately 1.097 x 10^7 m^-1)
- n1 is the initial energy level of the electron
- n2 is the final energy level of the electron
In the case of helium ion (He+), the ground state (n1) has an electron configuration of 1s^2, and the ion has one electron missing, resulting in a configuration of 1s^1.
We are looking for the smallest energy transition, which corresponds to the electron transitioning from a higher energy level to the lowest energy level (n2 = 1).
Substituting the values into the Rydberg formula, we can solve for the wavelength (λ):
1/λ = R * (1/n1^2 - 1/n2^2)
1/λ = (1.097 x 10^7 m^-1) * (1/1^2 - 1/1^2)
1/λ = (1.097 x 10^7 m^-1) * (1 - 1)
1/λ = 0
Since the denominator becomes 0, it implies that the wavelength is infinite (or undefined) for the smallest energy transition in helium ion (He+). This means that no emission occurs in the visible spectrum for this specific transition.