The ground stated energy for the hydrogen atom is 5.44 x 10-19J . If an electron drops from zero energy to ground state, calculate the frequency of the emit...
Question 1 Report
The ground stated energy for the hydrogen atom is 5.44 x 10-19J . If an electron drops from zero energy to ground state, calculate the frequency of the emitted radiation. [h = 6.6 x 10-34 Js]
Answer Details
When an electron drops from a higher energy level to a lower energy level, energy is released in the form of electromagnetic radiation (photons). The frequency of the radiation can be calculated using the formula:
E = hf
where E is the energy of the photon, h is Planck's constant, and f is the frequency of the radiation.
In this question, the energy of the photon is equal to the difference between the initial energy level (zero) and the final energy level (5.44 x 10^-19 J):
E = 5.44 x 10^-19 J
Planck's constant is given as h = 6.6 x 10^-34 Js. Substituting these values into the formula, we get:
5.44 x 10^-19 J = (6.6 x 10^-34 Js) x f
Solving for f, we get:
f = (5.44 x 10^-19 J) / (6.6 x 10^-34 Js)
f ≈ 8.24 x 10^14 Hz
Therefore, the frequency of the emitted radiation is approximately 8.24 x 10^14 Hz, which is not one of the given options.
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