Exciting an electron in a hydrogen atom.

[weston_sagle]

Homework Statement

A hydrogen atom in its ground state is illuminated with light having a wavelength of 96.7 nm. can the hydrogen atom's electron be excited to a higher energy state by absorbing one of these photons? If so, determine the state and wavelength of any photon that would be emitted by the atom.

Homework Equations

E=hc/lambda
En=-13.6eV/n^2

The Attempt at a Solution

E=hc/lambda
=[(6.626x10^-34)(3.0x10^8)/9.67x10-8]
=2.0556x10^18 J

Convert to / by 1.602x10^-19 eV = 12.83eV

En=-13.6eV/1^2 = -13.6eV

So I think that the photon energy would have to exceed 13.6eV to excite the electron, so in this case it would not excite it. Is this right?

 

[Nate810]

If it did excite the electron, would the wavelength of the photon emitted be longer or shorter than the one absorbed? Would the energy of the emitted photon be greater or less than that of the absorbed photon?

 

[tomcue]

Yes, your analysis is correct. In order to excite an electron in a hydrogen atom, the energy of the photon must be greater than the energy difference between the ground state and the excited state. In this case, the energy of the photon is 12.83eV, which is less than the ground state energy of -13.6eV. Therefore, the electron will not be excited by this photon.

To determine the state and wavelength of a photon that would excite the electron, we can use the equation En=-13.6eV/n^2. Plugging in the ground state energy of -13.6eV and solving for n, we get n=2. This means that the electron would be excited to the n=2 energy state.

To find the wavelength of the emitted photon, we can use the equation E=hc/lambda, where E is the energy difference between the n=2 and n=1 states, which is 10.2eV. Plugging in the values and solving for lambda, we get lambda=121.6nm. Therefore, the photon emitted by the excited hydrogen atom would have a wavelength of 121.6nm.