Quantum Physics: Electron Transition in Hydrogen Atom

Does the electron gain or lose energy during this transition?

Choose one:

a) Gain

b) Lose

Answer:

The electron loses 1.51 eV of energy during the transition.

During the transition of an electron from the n = 5 state to the n = 3 state in a hydrogen atom, the electron loses energy.

To determine the energy gained or lost, we can use the formula for the energy difference between two states in an atom: ΔE = E_final - E_initial. Substituting the values, ΔE = (−13.6eV/n_final^2) - (−13.6eV/n_initial^2). Plugging in n_final = 3 and n_initial = 5, we find ΔE = −1.51 eV.

Since the energy change is negative, the electron loses 1.51 eV of energy during this transition.

During this transition, does the atom absorb or emit a photon?

Choose one:

a) Absorb

b) Emit

Answer:

The atom emits a photon.

When an electron transitions between energy levels in an atom, it can absorb or emit a photon. In this case, since the electron is moving from a higher energy state to a lower energy state, the atom emits a photon.

What is the photon energy in Joule?

Answer:

The photon energy is also 1.51 eV.

What is the wavelength of the photon?

Answer:

The wavelength of the photon can be calculated using λ = c/f, where f is found by rearranging E_photon = hf and solving for f.

To find the wavelength of the photon, we can use the equation λ = c/f, where λ is the wavelength, c is the speed of light, and f is the frequency. Since we know the speed of light is 3 × 10^8 m/s, we need to find the frequency.

We can use the equation E_photon = hf, where h is Planck's constant. Rearranging the equation, we find f = E_photon/h. Plugging in the values, f = (1.51 eV)/(1.6 × 10^(-19) J/eV × 6.63 × 10^(-34) J·s), we can solve for f. Finally, using the equation λ = c/f, we can calculate the wavelength of the photon.