A hydrogen-like ion has a nucleus of charge +Ze and a single electron outside this nucleus. The energy levels of these ions are -Z2 RH/n2 (where Z is the atomic number). Calculate the wavelength of the transition (nm) from n = 3 to n = 2 for He+, a hydrogen- like ion. What would be the perceived colour of this emission?

Question

A hydrogen-like ion has a nucleus of charge +Ze and a single electron outside this nucleus. The energy levels of these ions are -Z2 RH/n2 (where Z is the atomic number). Calculate the wavelength of the transition (nm) from n = 3 to n = 2 for He+, a hydrogen- like ion. What would be the perceived colour of this emission?

Answer 1

To calculate the wavelength of the transition from n = 3 to n = 2 for He+, we can use the Rydberg formula for hydrogen-like ions:

1/λ = R(1/n2_final^2 - 1/n2_initial^2)

Where:
- λ is the wavelength of the transition,
- R is the Rydberg constant (approximately 1.097 × 10^7 m^-1),
- n2_final is the final energy level (n = 2),
- n2_initial is the initial energy level (n = 3).

First, let's substitute the values into the formula:

1/λ = R(1/2^2 - 1/3^2)

1/λ = R(1/4 - 1/9)

Now, let's calculate the fraction inside the parentheses:

1/λ = R(9/36 - 4/36)

1/λ = R(5/36)

Next, divide both sides of the equation by R:

1/λR = 5/36

Now, isolate λ on one side of the equation by taking the reciprocal of both sides:

λ = 36/5R

λ = (36/5)(1.097 × 10^7)^-1

Finally, calculate the value of λ:

λ ≈ 656.268 nm

Therefore, the wavelength of the transition from n = 3 to n = 2 for He+ is approximately 656.268 nm.

Now, to determine the color of this emission, we can use the concept of the visible light spectrum. The visible light spectrum ranges from approximately 400 nm (violet) to 700 nm (red). Since the calculated wavelength (656.268 nm) falls within this range, the perceived color of this emission would be red.