Radiation coming from transitions $n = 2$ to $n = 1$ of hydrogen atoms falls on helium ions in $n = 1$ and $n = 2$ states. What are the possible transition of helium ions as they absorb energy from the radiation ?

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Solution

By Bohr's model, energy in nth orbit is given by $\frac{- 13.6 Z^{2}}{n^{2}}$ where Z is the atomic number of the atom. Using the given formula, energies of different states are obtained as shown in the image. Now when transition is made from n=2 to n=1, a photon of 13.6-3.4=10.2 eV energy is released.

When this falls on helium atom:
When it falls on n=1 state, the minimum energy required to make a transition is 54.5-13.6=40.2 eV. Hence non transition is possible with this photon,
When it falls on n=2 state, a transition can directly be made from n=2 to n=4 as the energy required for this is 10.2 eV only. Or it can first make a transition to n=3 state and then n=4.

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Radiation coming from transitions n=2 to n=1 of hydrogen atoms falls on helium ions in n=1 and n=2 states. What are the possible transition of helium ions as they absorb energy from the radiation ?

Radiation coming from transitions $n = 2$ to $n = 1$ of hydrogen atoms falls on helium ions in $n = 1$ and $n = 2$ states. What are the possible transition of helium ions as they absorb energy from the radiation ?