Question

Energy E of a hydrogen atom with principal quantum number n is given by E = \(-\frac{13.6}{n^2}eV\). The energy of a photon ejected when the electron jumps from the n = 3 state to the n = 2 states of hydrogen is approximately

• A. 0. 85 eV
• B. 3.4 eV
• C. 1.9 eV
• D. 1.5 eV

Answer 1

The Correct Option is C

To find the energy of the photon ejected when an electron in a hydrogen atom transitions from the \(n = 3\) state to the \(n = 2\) state, we need to calculate the energy difference between these two states using the given formula for the energy of hydrogen atom energy levels:

The energy \(E_n\) of an electron in a hydrogen atom at the principal quantum number \(n\) is given by:

E_n = -\frac{13.6}{n^2} \,\text{eV}

First, calculate the energy for \(n = 3\):

E_3 = -\frac{13.6}{3^2} = -\frac{13.6}{9} \approx -1.51 \,\text{eV}

Next, calculate the energy for \(n = 2\):

E_2 = -\frac{13.6}{2^2} = -\frac{13.6}{4} = -3.4 \,\text{eV}

The energy of the photon ejected, which is equal to the energy difference between these two states, is:

\Delta E = E_2 - E_3 = -3.4 \,\text{eV} - (-1.51 \,\text{eV}) = -3.4 + 1.51\approx -1.89 \,\text{eV}

The magnitude of this energy is approximately \(1.9 \,\text{eV}\) as energy is always expressed as a positive quantity for photons.

Therefore, the energy of the photon ejected is approximately \(1.9 \,\text{eV}\).

Conclusion: The correct answer is 1.9 eV.