determine the wavelength of the line in the emission spectrum produced by an electron transition He+ from ni= 4 to nf= 2.
1/wavelength = RZ^2(1/2^2 - 1/4^2)
R is the Rydberg constant which is 1.09737E7 and Z = 2(don't forget to square it). You can find R more accurately if you go to google.com and type in rydberg constant and click on the wikipedia link.
To determine the wavelength of the line in the emission spectrum produced by an electron transition from ni = 4 to nf = 2 in He+, we can use the Rydberg formula and the Balmer series equation.
The Rydberg formula calculates the wavelength of light emitted or absorbed during an electron transition in an atom:
1/λ = R * (1/ni^2 - 1/nf^2)
Where λ is the wavelength of light, R is the Rydberg constant (approximately 1.097 × 10^7 m^-1), ni is the initial energy level, and nf is the final energy level.
For the Balmer series of hydrogen-like atoms (including He+), the formula simplifies as follows:
1/λ = R * (1/4 - 1/2)
Substituting the values:
1/λ = R * [1/4 - 1/2]
1/λ = R * [1/4 - 2/4]
1/λ = R * [-1/4]
Now, let's solve for λ, the wavelength:
1/λ = -R/4
Multiplying both sides by λ:
1 = - Rλ/4
Rearranging the equation:
λ = -4 / (R)
Now substitute the value of R = 1.097 × 10^7 m^-1 to get the final result:
λ = -4 / (1.097 × 10^7 m^-1)
Calculate the value:
λ ≈ -3.64 × 10^-8 meters
Since wavelength cannot be negative, we take the absolute value:
λ ≈ 3.64 × 10^-8 meters
Therefore, the wavelength of the line in the emission spectrum produced by the electron transition from ni = 4 to nf = 2 in He+ is approximately 3.64 × 10^-8 meters.