Question

3.5. The n=1 electron energy level of a H

atom has an energy of -21.8 x10-¹9 J. What is
the energy of the n=2 level?
3.6. Using the result from the problem above,
calculate the energy of a photon emitted when
an electron jumps down from n=2 to n=1 in a
H atom.

Answer 1

Explanation :

The energy of an electron in the n=2 energy level can be calculated using the energy equation for hydrogen atoms:

E = -13.6 eV/n²

where E is the energy in electron volts (eV) and n is the principal quantum number. Converting the energy given in the problem from joules to electron volts:

-21.8 x 10^(-19) J = -21.8 x 10^(-19) J x (6.242 x 10^18 eV/J) ≈ -136.1 eV

Plugging in n=2 into the energy equation:

E = -13.6 eV/2²

E = -13.6 eV/4

E ≈ -3.4 eV

Therefore, the energy of the n=2 energy level is approximately -3.4 eV.

When an electron jumps down from n=2 to n=1, it emits a photon with energy equal to the difference in energy between the two energy levels. Using the energy values calculated above:

∆E = (-3.4 eV) - (-136.1 eV)

∆E = 132.7 eV

The energy of the photon emitted is approximately 132.7 electron volts (eV).