A hydrogen atom is initially in state with n = 1. Can it emit a photon with a wavelength in the visible part of the electromagnetic spectrum?
The correct answer and explanation is:
Answer:
No, a hydrogen atom initially in the ground state (n = 1) cannot emit a photon with a wavelength in the visible part of the electromagnetic spectrum.
Explanation (300 words):
The hydrogen atom’s energy levels are quantized, meaning electrons can only exist in certain discrete energy states labeled by the principal quantum number nn. The energy EnE_n of the hydrogen atom at level nn is given by the formula: En=−13.6 eVn2E_n = -\frac{13.6 \text{ eV}}{n^2}
where 13.6 eV is the ionization energy of hydrogen at the ground state n=1n=1.
Photon Emission and Transitions:
An electron emits a photon when it transitions from a higher energy level nin_i to a lower energy level nfn_f. The energy of the emitted photon corresponds to the energy difference between these levels: Ephoton=Eni−EnfE_{\text{photon}} = E_{n_i} – E_{n_f}
The wavelength λ\lambda of the emitted photon relates to its energy by: Ephoton=hcλE_{\text{photon}} = \frac{hc}{\lambda}
where hh is Planck’s constant and cc is the speed of light.
Initial State n=1n=1:
Since the electron starts in the ground state n=1n=1, it cannot emit a photon by dropping to a lower level because there is no lower level than the ground state. For emission, the electron must first be excited to a higher energy level n>1n > 1, then fall back down to a lower energy level, releasing photons of specific wavelengths.
Visible Spectrum and Hydrogen Lines:
The visible spectrum ranges approximately from 400 nm (violet) to 700 nm (red). In hydrogen, visible photons are emitted when electrons fall to the n=2n=2 level from higher levels. This set of transitions forms the Balmer series, which includes wavelengths such as 656 nm (red), 486 nm (blue-green), and others in the visible range.
If the atom starts in n=1n=1, the electron must first be excited to a higher level (n≥3)(n \geq 3), then drop to n=2n=2 to emit visible light. Direct transitions to n=1n=1 from higher levels emit photons in the ultraviolet (Lyman series), not visible light.
Summary:
- Electron initially in n=1n=1 can’t emit photons by dropping to a lower state because n=1n=1 is the lowest state.
- Visible photons are emitted during transitions to n=2n=2, which require the electron to be in a higher excited state first.
- Direct emissions from excited states down to n=1n=1 produce ultraviolet photons, not visible photons.
Therefore, a hydrogen atom initially in n=1n=1 cannot directly emit visible light photons.