Investigating Helium Ion's Wavelength Transition

[matpo39]

here is the question I am stuck on:
Radiation from a helium ion He+ is nearly equal to the wavelength to the $$H_\alpha$$ line (the first line of the Balmer series). (a) Between what states (values of n) does the transition in the helium ion occur? (b) is the wavelenght greater or smaller than that of the $$H_\alpha$$ line?

my first attempt to solve this was to use the eqation

$$\frac{1}{\lambda}=R(\frac{1}{n_f^2}-\frac{1}{n_i^2})$$

and i would set $$\lambda = 6562.8 \AA$$the wave length of $$H_\alpha$$ but i still can't solve for because both of the states are unknown. anyone have any suggestions on this?

thanks

 

[quantumdude]

You know that $n_f$ and $n_i$ both have to be natural numbers (1,2,3,...). You can start by trying combinations.

 

[quicknote]

I would approach this question by first understanding the concept of wavelength transition in atoms. Wavelength transition occurs when an electron in an atom moves from a higher energy state to a lower energy state, emitting a photon of light with a specific wavelength. The energy states of an atom are described by the quantum numbers n, l, and m, where n represents the principal energy level, l represents the orbital angular momentum, and m represents the magnetic quantum number.

In the case of a helium ion He+, the transition is between the n=2 and n=1 states. This can be determined by looking at the energy level diagram for a helium atom, which shows that the H_\alpha line corresponds to the transition from the n=2 to n=1 state.

To determine whether the wavelength is greater or smaller than that of the H_\alpha line, we can use the Rydberg formula that you mentioned. However, we need to rearrange the equation to solve for n_f (final energy state) instead of lambda (wavelength).

n_f=\sqrt{\frac{1}{1-\frac{\lambda}{R}}}

We know that the final energy state n_f=1 for the H_\alpha line, so we can plug in the wavelength of 6562.8 Å and solve for n_i (initial energy state).

n_i=\sqrt{\frac{1}{1-\frac{6562.8 \AA}{R}}}=2

This means that the wavelength of the helium ion's transition is smaller than that of the H_\alpha line, as it occurs between the n=2 and n=1 states instead of n=2 and n=1.

In conclusion, the transition in a helium ion He+ occurs between the n=2 and n=1 states, and the wavelength is smaller than that of the H_\alpha line. This information can be used for further investigation and understanding of the properties and behavior of helium ions.