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?
Are you sure about that formula you gave for the energy level? For some units that COULD be right. As it stands I don't that it is correct. I think it should be E = -hcRZ^2(1/n^2). Whatever formula you use, here is what you do. Calculate the E for n = 2 and E for n = 3. Z for He^+ = 2. Get the difference between E3 and E2(it should be positive) if you subtract correctly. Then delta E = hc/w where h is Planck's constant, c is speed of light in m/s and w is wavelength in m/s. You want the answer in nm. Convert remembering that 1 m = 10^9 nm. Post your work if you get stuck but check out that formula
1.64
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 = RH (1/n2 - 1/n1)
where λ is the wavelength, RH is the Rydberg constant, and n1 and n2 are the initial and final energy levels, respectively.
For He+, the atomic number Z is 2.
Plugging in the values:
1/λ = RH (1/2^2 - 1/3^2)
Simplifying:
1/λ = RH (1/4 - 1/9)
1/λ = RH (9/36 - 4/36)
1/λ = RH (5/36)
Now, we need the value of the Rydberg constant, which is given as RH = 1.097 x 10^7 m^-1.
1/λ = (1.097 x 10^7 m^-1)(5/36)
Simplifying further:
1/λ = 1527777.778 / m
Now, find the reciprocal of both sides to solve for λ:
λ = 1 / (1527777.778 / m)
λ = m / 1527777.778
To convert meters to nanometers, multiply by 10^9:
λ (nm) = (m / 1527777.778) x 10^9
Substituting the value of m:
λ (nm) ≈ (3.07 x 10^-7 m) / 1527777.778
Calculating:
λ (nm) ≈ 201.4 nm
The calculated wavelength of the transition from n = 3 to n = 2 for He+ is approximately 201.4 nm.
To determine the perceived color of this emission, we can use the visible light spectrum. Light with shorter wavelengths appears more towards the violet end, while light with longer wavelengths appears more towards the red end.
In this case, the calculated wavelength of 201.4 nm falls in the range of the ultraviolet (UV) spectrum, which is invisible to the human eye. Therefore, the perceived color of this emission would be ultraviolet.
To calculate the wavelength of the transition from n = 3 to n = 2 for He+, we can use the Rydberg formula:
1/λ = R(1/n1^2 - 1/n2^2)
where λ is the wavelength, R is the Rydberg constant, n1 is the initial energy level, and n2 is the final energy level.
In this case, we have He+ as a hydrogen-like ion, so Z (the atomic number) is equal to 2. The Rydberg constant, RH, for hydrogen-like ions is the same as for hydrogen, so we can simply use RH.
Plugging in the numbers:
1/λ = RH(1/2^2 - 1/3^2)
Simplifying:
1/λ = RH(1/4 - 1/9)
1/λ = RH(9/36 - 4/36)
1/λ = RH(5/36)
Now we can substitute the value of RH, which is 1.097 x 10^7 m^-1, to find the value of 1/λ:
1/λ = (1.097 x 10^7 m^-1) * (5/36)
Multiplying:
1/λ = 1.524 x 10^6 m^-1
Now we can calculate the value of λ by taking the reciprocal of both sides:
λ = 1 / (1.524 x 10^6 m^-1)
λ = 6.56 x 10^-7 m
Finally, to convert the wavelength to nanometers (nm):
λ (in nm) = (6.56 x 10^-7 m) * (10^9 nm/m)
λ (in nm) ≈ 656 nm
So the wavelength of the transition from n = 3 to n = 2 for He+ is approximately 656 nm.
To determine the perceived color of the emission, we can refer to the visible light spectrum. Colors towards the higher end of the spectrum have shorter wavelengths, while colors towards the lower end have longer wavelengths.
656 nm falls within the red range of the visible light spectrum. Therefore, the perceived color of the emission would be reddish.