In the Sun, an ionized helium atom makes a transition from n =6 state to the n = 2 state, emitting a photon. Can that photon be absorbed by hydrogen atoms present in the Sun? If so between what energy states will the hydrogen atom jump?

It seems like I'm answering this 5 years later, but this is for anyone who needs this in the future!

Helium is a hydrogen-like atom. Using the fact that the ground state energy of hydrogen is 13.6eV, we can use the relation:

E = 13.6eV(z^2/n^2) to calculate the energy levels of helium, where z is charge number and n^2 is the principle quantum number.

For Helium z = 2, and in our case n1 = 6 and n2 = 2.

Therefore,

Change in E = 13.6eV*z^2* (1/(n2^2) - 1/(n1^2))

That is,
Change in E = 13.6*4*((1/4)-(1/36)) = 12.1eV.

In words, the ionised helium atom's photon emission requires 12.1eV, and that is the energy of the photon.

Now, this 12.1eV must be compared with hydrogen state transition energies.

For H:
Energy(in a given state) = -13.6eV(1/(n^2))

I will replace n by m here to avoid confusion, m for hydrogen where n was used for helium.

By comparing changes in energy between levels, where m1 = 1 and m2 = 3 yields a change in energy = 12.1eV.

So a hydrogen atom will absorb the photon and experience a transition, as above.

Well, well, well, let's unravel this mystery, shall we? So, we have an ionized helium atom in the Sun, getting all excited and jumping from the n = 6 state to the n = 2 state, and it releases a photon in the process. Now, if there are hydrogen atoms loitering around in the Sun, could they possibly snatch this photon for themselves?

The answer is, unfortunately for the hydrogen atoms, no comedy show here today. You see, for the hydrogen atom to absorb a photon, it needs the energy of the photon to match the energy difference between its own energy levels. In this case, the photon emitted by the helium atom during its transition from n = 6 to n = 2 has an energy corresponding to that specific transition.

However, the hydrogen atom won't be able to absorb this photon because the energy difference between its energy levels doesn't match with the energy of the helium photon. It's like trying to fit a king-size bed into a clown car. Just not gonna happen, my friend.

So, to answer your second question, the hydrogen atoms won't make any jumps because they won't be absorbing this particular photon. They'll simply sit there, minding their own business, waiting for a more suitable photon to come along and brighten their day.

Hope that clears things up for you!

Yes, the photon emitted by the ionized helium atom can be absorbed by hydrogen atoms present in the Sun. To determine between what energy states the hydrogen atom will jump, we need to consider the energy difference between the initial and final states.

The energy difference between two energy states in an atom is given by the formula:

ΔE = E(final state) - E(initial state)

For a hydrogen atom, the energy of an electron in the nth energy level is given by the equation:

E(n) = -13.6 eV / n^2

Using this equation, we can calculate the energy difference between the n = 6 and n = 2 states in the helium atom:

ΔE(He) = E(n = 2) - E(n = 6)
= (-13.6 eV / 2^2) - (-13.6 eV / 6^2)
= -13.6 eV / 4 + 13.6 eV / 36
= -3.4 eV + 0.3777778 eV
= -2.7722222 eV

Now, to find the energy levels of the hydrogen atom that correspond to this energy difference, we can equate it to the energy difference between two energy levels in a hydrogen atom:

ΔE(H) = E(final state) - E(initial state)
= -13.6 eV / n^2 - E(initial state)

-2.7722222 eV = -13.6 eV / n^2 - E(initial state)

Let's solve for n^2 using this equation:

-2.7722222 eV + E(initial state) = -13.6 eV / n^2

n^2 = (-13.6 eV) / (-2.7722222 eV + E(initial state))

The values of n^2 that satisfy this equation will correspond to the possible energy states that the hydrogen atom can jump to upon absorbing the photon.

To determine whether the photon emitted by the ionized helium atom can be absorbed by hydrogen atoms in the Sun and between what energy states the hydrogen atom will jump, we need to compare the energy difference of the emitted photon (initial transition) with the energy levels of the hydrogen atom.

First, let's find the energy difference between the n = 6 and n = 2 states for the ionized helium atom. The energy difference can be calculated using the formula:

ΔE = E_final - E_initial

Given that the initial state (n = 6) and final state (n = 2), we can use the formula for calculating the energy of an electron in a hydrogen-like atom:

E = -13.6 eV / n^2

Using this formula, we can calculate the initial energy (E_initial) and the final energy (E_final) for the helium atom:

E_initial = -13.6 eV / 6^2 = -0.378 eV
E_final = -13.6 eV / 2^2 = -3.4 eV

Now, we need to determine whether the energy of the photon emitted by the helium atom falls within the energy levels of the hydrogen atom. The difference in energy between the initial and final states of the helium atom corresponds to the energy of the emitted photon.

The energy of a photon is given by the equation:

E_photon = h * ν

Where E_photon is the energy of the photon, h is Planck's constant (6.63 x 10^-34 J·s), and ν is the frequency of the photon.

To calculate the energy of the photon in electron volts (eV), we can use the conversion factor 1 eV = 1.6 x 10^-19 J. Hence:

E_photon (in eV) = (h * ν) / (1.6 x 10^-19)

As we know the energy of the photon, we can compare it to the energy states of the hydrogen atom to determine if the photon can be absorbed.

The energy levels of a hydrogen atom are given by the equation:

E = -13.6 eV / n^2

We need to check which energy level is closest to the energy of the emitted photon. By comparing the energy levels, we can determine if the photon can be absorbed by the hydrogen atom and between what energy states the hydrogen atom will transition.

To summarize, the process involves:
1. Calculating the energy difference of the ionized helium atom between the initial and final states.
2. Determining the energy of the emitted photon.
3. Comparing the energy of the photon with the energy levels of the hydrogen atom to see if it can be absorbed.
4. Identifying the closest energy levels of the hydrogen atom to determine the jump in energy states.

Performing these calculations will give us the answer to whether the photon can be absorbed by hydrogen atoms in the Sun and the corresponding energy states of the hydrogen atom transition.