From the energy-level diagram for hydrogen, explain the observation that the frequency of the second Lyman-series line is the sum of the frequencies of the first Lyman-series line and the first Balmer-series line. This is an example of the empirically discovered Ritz combination principle. Use the diagram to find some other valid combinations.
It is shown that how the frequency of the second Lyman-series line is the sum of the frequencies of the first Lyman-series line and the first Balmer-series.
Another example of such transition is the transition associated with the second Balmer-series line can be thought of as (first Paschen) followed by (first Balmer).
The primary quantum number, " n," has a direct impact on the energy level diagram. To calculate the energy difference between two states and describe the hydrogen spectrum, an energy diagram is required.
The energy level diagram of the hydrogen atom is displayed in the diagram below.
Let the transition is from to
So, the change in energy is given by,
This decrease in energy emerges in the form of radiation, having energy hv , such that-
Here, v is the frequency of photon released during transition and h is Planck’s constant, whose value is .
Now, the transition can be done in another way also.
Which is to
So, by equation (1) change in energy for to is given by,
Here, we have two transitions taking place, so two photons having energies are released having frequencies role="math" localid="1661856051342" respectively. Also, we have-
Now, by equating equation (1), (2) and (3) we get,
So, the transition associated with the second Lyman-series line can be thought of as (first Balmer) followed by (first Lyman). From the energy level diagram, we have-
These values are related as-
Hence, the frequency of second Lyman-series line is the sum of the frequency of first Balmer series line and the first Lyman-series line.
From the energy level diagram, we can observe that the transition associated with the second Balmer-series line can be thought of as (first Paschen line) followed by (first Balmer line).
In atoms, there is a finite, though very small, probability that, at some instant, an orbital electron will actually be found inside the nucleus. In fact, some unstable nuclei use this occasional appearance of the electron to decay by electron capture. Assuming that the proton itself is a sphere of radius and that the wave function of the hydrogen atom’s electron holds all the way to the proton’s center, use the ground-state wave function to calculate the probability that the hydrogen atom’s electron is inside its nucleus.
An electron is trapped in a one-dimensional infinite potential well. For what (a) higher quantum number and (b) lower quantum number is the corresponding energy difference equal to the energy of the n = 5 level? (c) Show that no pair of adjacent levels has an energy difference equal to the energy of the n = 6 level.
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