Calculate the frequency of light associated with the emission from n=5 to n=4 in the hydrogen atom.
I got -7.41 x10^13 Hz, but the answer would make sense if it is positive... So does this mean that when the electrons jump from n=5 to n=4, energy is emitted, but the energy has to change to positive??
Is the answer then
+7.41 x 10^13 Hz?
E = R(1/16 - 1/25) and that is a positive number. You probably got a negative number by making it 1/25 - 1/16.
To calculate the frequency of light associated with the emission from n=5 to n=4 in the hydrogen atom, you can use the Rydberg formula. The Rydberg formula relates the frequency of light emitted to the energy difference between two energy levels.
The formula is:
1/λ = R_H * (1/n_final^2 - 1/n_initial^2)
where λ is the wavelength of light emitted, R_H is the Rydberg constant for hydrogen (approximately 3.29 x 10^15 Hz), and n_final and n_initial are the final and initial energy levels, respectively.
In this case, the initial energy level is n=5 and the final energy level is n=4. Plugging these values into the formula, you get:
1/λ = R_H * (1/4^2 - 1/5^2)
Simplifying further:
1/λ = R_H * (1/16 - 1/25)
1/λ = R_H * (25 - 16) / 400
1/λ = 9R_H / 400
Now, let's substitute the value of R_H, which is approximately 3.29 x 10^15 Hz:
1/λ = (9 * 3.29 x 10^15 Hz) / 400
1/λ ≈ 2.98 x 10^13 Hz
To find the frequency, we can take the reciprocal of λ:
λ ≈ 3.36 x 10^-14 meters
Therefore, the frequency of light associated with the emission from n=5 to n=4 in the hydrogen atom is approximately 2.98 x 10^13 Hz.
Note: The negative sign in your initial answer (-7.41 x 10^13 Hz) might be due to an error in the calculation or in the use of the Rydberg formula. The energy difference between n=5 and n=4 is a positive value, indicating that energy is emitted when the electron transitions from a higher energy level to a lower energy level.