Suppose an e- in an H atom has a transition from n = 3 to n = 2.
a. Determine the energy of the released photon.
i set it up like this
Ei=-Rh/3^2 Ef=-Rh/2^2
(-Rh/9)-(-rh/4)=-4rh+9rh/36=/-rh+4rh/9/- this part why does it go to -rh+4rh? But then I got the end result which is 3rh/9 =hv
Then I did 3/9 x 2.179x10^-18 j/6.626 x 10^-34 j.s =1.10 x 10^17/s
b. determine what type of light is emitted.
Show all units and conversion factors.
For this part I did 3.00x10^8 m/s/1.10x10^-6m =273 nam
Is this correct?
I follow some of what you did but not all of it and I don't get the same answer you did. Why not use
dE = 2.180E-18J(1/4-1/9)
Then lambda = hc/dE
For dE I have about 3.03E-19J
and wavelength = about 6.56E-7 m or 656 nm or 6560 angstroms which is red.
Let's break down your calculations step by step to see where the confusion arises.
a. Determining the energy of the released photon:
You correctly set up the initial and final energy levels for the transition as follows:
Ei = -Rh/3^2 <-- Initial energy level
Ef = -Rh/2^2 <-- Final energy level
The equation you used to calculate the energy difference is:
ΔE = Ef - Ei = -Rh/4 - (-Rh/9) = -Rh/4 + Rh/9
To simplify this expression, we need to find a common denominator for the fractions -Rh/4 and Rh/9. The common denominator is 36:
ΔE = (-Rh/4)*(9/9) - (Rh/9)*(4/4)
= -9Rh/36 + 4Rh/36
= (-9Rh + 4Rh)/36
= -5Rh/36
So, the energy difference (ΔE) between the two energy levels is -5Rh/36.
Next, you correctly substitute the value for Rh and simplify:
ΔE = -5Rh/36 = -5*(2.179x10^-18)/36
≈ -2.899x10^-19 J
Since the energy of the photon is equal to the absolute value of the energy difference, the energy of the released photon is approximately 2.899x10^-19 J.
b. Determining the type of light emitted:
To determine the type of light emitted, we need to find the wavelength (λ) of the photon using the energy of the photon (E) and the speed of light (c):
E = hc/λ
Rearranging the equation, we get:
λ = hc/E
Substituting the known values:
λ = (6.626x10^-34 J.s * 3.00x10^8 m/s) / (2.899x10^-19 J)
≈ 6.845x10^-7 m
So, the wavelength of the emitted light is approximately 6.845x10^-7 meters, which corresponds to the red region of the visible light spectrum.
Please note that there might be slight differences in numerical values due to rounding during calculations.