Determine the energy and n(i) for the spectral line at 121.6 nm (orange). Hint: R(H)=2.18x10^-18 J, c=3.0x10^8, and 1x10^9 nm - 1 m.
Line, nm
121.6
E, J
___
n(f)
1
n(i)
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I can't tell what you're doing but note that the Rydberg constant for H is NOT the number you have listed.
To determine the energy and n(i) for the spectral line at 121.6 nm (orange), we can use the Rydberg formula and the conversion factors provided.
The Rydberg formula for the spectral lines of hydrogen is given by:
1/λ = R(H) * (1/n(i)^2 - 1/n(f)^2)
where:
- λ is the wavelength of the spectral line
- R(H) is the Rydberg constant for hydrogen (2.18x10^-18 J)
- n(i) is the initial energy level (or initial quantum number)
- n(f) is the final energy level (or final quantum number)
In this case, we have the wavelength (λ) of 121.6 nm for the orange spectral line.
First, we need to convert the wavelength from nanometers (nm) to meters (m). We can use the conversion factor provided: 1x10^9 nm = 1 m.
121.6 nm = 121.6 x (1x10^-9 m/1 nm) = 121.6 x 10^-9 m
Now, we can substitute the values into the Rydberg formula:
1/(121.6 x 10^-9 m) = 2.18x10^-18 J * (1/n(i)^2 - 1/n(f)^2)
To solve for n(i), we rearrange the equation:
1/n(i)^2 = 1/(2.18x10^-18 J * (1/n(f)^2) + 1/(121.6 x 10^-9 m))
Now, plug in the values:
1/n(i)^2 = 1/(2.18x10^-18 J * (1/1^2) + 1/(121.6 x 10^-9 m))
Simplifying:
1/n(i)^2 = 1/(2.18x10^-18 J + 1/(121.6 x 10^-9 m))
Now, calculate the sum in the denominator:
1/n(i)^2 = 1/(2.18x10^-18 J + 1/(121.6 x 10^-9 m))
= 1/(2.18x10^-18 J + 1/(121.6 x 10^-9 m))
≈ 1/(2.18x10^-18 J) (Neglecting the small term 1/(121.6 x 10^-9 m))
Finally, solve for n(i):
1/n(i)^2 ≈ 1/(2.18x10^-18 J)
Taking the reciprocal:
n(i)^2 ≈ 2.18x10^-18 J
Solving for n(i):
n(i) ≈ √(2.18x10^-18 J)
Using a calculator, evaluate the square root:
n(i) ≈ 1
Therefore, the energy of the spectral line at 121.6 nm is approximately 2.18x10^-18 J, and the initial energy level (n(i)) is approximately 1.