Question

What is wavelength of a photon that would be emitted when a hydrogen atom makes a transition from the n = 3 to the n = 2 energy level? What is the longest wavelength photon that could ionize a hydrogen atom originally in its ground state? What processes might occur if an electron of energy 12.2 eV collides with a hydrogen atom at rest and in its ground state?

Answer 1

A.) The relation between different transitions of Hydrogen atom and the corresponding wavelength of the photon emitted is 1/λ = R_H [1/n₁²   - 1/n₂²  ] where R_H is Rydberg's constant whose value is 1.0973731 x 10⁷ m^-1

when the transition is from n₂ to n₁  

so putting n₂ =3 and n₁ =2 , we get   

1/λ = 1.0973731 x 10⁷ [ 1/2²  - 1/3² ]

λ = 6.561123104 x 10^-7 m ~ 656 nm

B.) Given that, the atom is originally in the ground state. So, n₁ =1 . For, longest wavelength the energy corresponding should be minimum. Which means if n₁ =1 is fixed then n₂ should be 2 for minimum energy and hence maximum wavelength.

putting n₂ = 2 and n₁ =1 , we get

1/λ = 1.0973731 x 10⁷ [ 1/1² - 1/2² ]

λ = 1.215022797 x 10^-7 m ~ 121 nm

C.) The same expression for various transitions of hydrogen atom written in terms of energies looks like this

hν = E₂ - E₁ = -13.6 [ 1/n₁²  - 1/ n₂²  ] eV

Also given that the hydrogen atom is in its ground state which means n₁ = 1 .

putting n₂ = 3 gives - 12.1 eV

which is the closest to the given number (12.2 eV).

Under any circumstances the colliding electron is NOT going to go to one of the energy levels of the atom but, if it can transfer/ deposit 12.1 eV of its energy to the already existing electron in the ground state, then the electron in the ground state might get excited to the third level (n = 3) from the ground state (n = 1). But still, the question says 12.2 eV and not 12.1 eV. So, I would say that nothing really happens here.