Caculate the wavelength of light emitted when each of the following transitions occur in the hydrogen atom.
A.) n=3->n=2
B.) n=4->n=3
C.) n=2->n=1
How do you do this. Could you please just show me step by step for atleaset one. Please!
1/wavelength = R*(1/n1^2 - 1/n2^2)
R = 1.09737E7
I tried to write n1 and square the n, not square the 1. Likewise, n2 and square the n and not the 2. The numbers are
1/n1^2 = 1/2^2 = 1/4
1/n2^2 = 1/3^2 = 1/9
wavelength in meters.
1/w =1.09737E7(1/4 - 1/9)
Solve for w.
Certainly! I can help you with one of the transitions step by step.
Let's calculate the wavelength of light emitted when the transition from n=4 to n=3 occurs in the hydrogen atom.
Step 1: Find the energy difference between the two energy levels.
The formula to calculate the energy difference (ΔE) is given by:
ΔE = Rh * (1/n2^2 - 1/n1^2)
where Rh is the Rydberg constant (approximately 2.18 x 10^-18 J), n1 is the initial principal quantum number (4 in this case), and n2 is the final principal quantum number (3 in this case).
ΔE = (2.18 x 10^-18 J) * (1/3^2 - 1/4^2)
= (2.18 x 10^-18 J) * (1/9 - 1/16)
= (2.18 x 10^-18 J) * (7/144)
= 1.068 x 10^-19 J
Step 2: Convert the energy difference to wavelength using the formula:
E = hc/λ
where E is the energy difference (in Joules), h is Planck's constant (approximately 6.63 x 10^-34 J·s), c is the speed of light (approximately 3.00 x 10^8 m/s), and λ is the wavelength (in meters).
Rearranging the equation, we have:
λ = hc/E
λ = (6.63 x 10^-34 J·s * 3.00 x 10^8 m/s) / (1.068 x 10^-19 J)
= (1.989 x 10^-25 J·m) / (1.068 x 10^-19 J)
= 1.866 x 10^-6 m
So the wavelength of light emitted when the transition from n=4 to n=3 occurs in the hydrogen atom is approximately 1.866 x 10^-6 meters.
You can follow the same steps to calculate the wavelengths for the other transitions given in the question.