Calculating energy and frequency of radiation emitted by an electron jump in a hydrogen atom

To calculate the energy and frequency of radiation emitted when an electron jumps from n=3 to n=2 in a hydrogen atom, we need to use the formula for the energy of a hydrogen atom:

E = -13.6 eV/n^2

where E is the energy of the electron, n is the quantum number of the electron shell, and eV is the electron volt.

Step 1: Calculating the energy of the electron in the initial state

Using the above formula, we can calculate the energy of the electron in the initial state (n=3):

E3 = -13.6 eV/3^2 = -1.51 eV

Step 2: Calculating the energy of the electron in the final state

Similarly, we can calculate the energy of the electron in the final state (n=2):

E2 = -13.6 eV/2^2 = -3.4 eV

Step 3: Calculating the energy of the emitted radiation

The energy of the emitted radiation can be calculated by taking the difference between the energy of the electron in the initial and final states:

ΔE = E3 - E2 = 1.89 eV

Step 4: Calculating the frequency of the emitted radiation

We can use the formula for the energy of a photon to calculate the frequency of the emitted radiation:

E = hf

where E is the energy of the photon, h is Planck's constant, and f is the frequency of the photon.

Rearranging the formula, we get:

f = E/h

Substituting the value of ΔE from step 3 and the value of Planck's constant (h = 6.626 x 10^-34 J.s), we get:

f = ΔE/h = (1.89 x 1.6 x 10^-19 J)/(6.626 x 10^-34 J.s) = 4.57 x 10^14 Hz

Therefore, the frequency of the radiation emitted by the electron jump from n=3 to n=2 in a hydrogen atom is 4.57 x 10^14 Hz.

Conclusion

In summary, the energy and frequency of the radiation emitted when an electron jumps from n=3 to n=2 in a hydrogen atom are 1.89 eV and 4.57 x 10^14 Hz, respectively. This calculation is based on the energy formula for a hydrogen atom and the formula for the energy of a photon.