An isolated atom of a certain gas has three relevant electron energy levels: E1=-5.60 eV (the ground state), E2=-3.10 eV, and E3=-1.20 eV. (a) Explain what is meant by stating that the electron energies of the atom are quantised. [1] (b) An electron in the atom makes a transitio…

Answer: ((a)) the atom can only have certain discrete (fixed) values of internal energy, not a continuous range of values ((b)) E=4.40 eV=7.04×10^-19 J ((c)) f=1.06×10^15 Hz, =283 nm, in the ultraviolet region ((d)) only photons whose energy exactly matches an energy-level gap of the atom are absorbed by the ground-state atoms; this removes those specific wavelengths from the transmitted beam, producing dark lines at those wavelengths. Part (a): Quantised energy levels Stating that the atom's electron energies are quantised means the atom can only exist with certain fixed, discrete values of internal energy (here, only E1, E2 or E3) and no values in between. The electron…

A Level Physics 9702: Atomic Energy Levels, Line Spectra and Photon Emission

An isolated atom of a certain gas has three relevant electron energy levels: $E_1=-5.60\text{ eV}$ (the ground state), $E_2=-3.10\text{ eV}$ , and $E_3=-1.20\text{ eV}$ .

(a) Explain what is meant by stating that the electron energies of the atom are quantised. [1]

(b) An electron in the atom makes a transition from level $E_3$ to level $E_1$ , emitting a photon. Calculate the energy of the emitted photon, giving your answer in both electron-volts and joules. [2]

(c) Calculate the frequency and the wavelength of this emitted photon, and state which region of the electromagnetic spectrum it lies in. [3]

(d) The atom is now in its ground state $E_1$ . A beam of white light, containing a continuous range of wavelengths, is passed through a sample of this gas. Explain, in terms of photon absorption, how dark absorption lines can appear in the transmitted spectrum. [2]

Part (a): Quantised energy levels

Stating that the atom’s electron energies are quantised means the atom can only exist with certain fixed, discrete values of internal energy (here, only $E_1$ , $E_2$ or $E_3$ ) and no values in between. The electron cannot have any arbitrary energy; it can only occupy one of these specific allowed levels.

Part (b): Energy of the emitted photon

When an electron falls from a higher energy level to a lower one, the atom loses energy equal to the difference between the two levels, and this energy is carried away as a single photon: $\Delta E = E_3 - E_1$

$\Delta E = -1.20 - (-5.60) = 4.40\text{ eV}$

Converting to joules: $\Delta E = 4.40\times1.60\times10^{-19} = 7.04\times10^{-19}\text{ J}$

Recompute as a check: $7.04\times10^{-19}/1.60\times10^{-19}=4.40\text{ eV}$ , which matches the starting value in eV.

Part (c): Frequency, wavelength and spectral region

The photon energy is related to frequency by $\Delta E = hf$ , so: $f = \frac{\Delta E}{h} = \frac{7.04\times10^{-19}}{6.63\times10^{-34}} = 1.06\times10^{15}\text{ Hz}$

The wavelength follows from $c=f\lambda$ : $\lambda = \frac{c}{f} = \frac{3.00\times10^{8}}{1.06\times10^{15}} = 2.83\times10^{-7}\text{ m} = 283\text{ nm}$

Recompute as a check, using the combined formula $\lambda=hc/\Delta E$ directly: $\lambda=\dfrac{6.63\times10^{-34}\times3.00\times10^{8}}{7.04\times10^{-19}}=\dfrac{1.989\times10^{-25}}{7.04\times10^{-19}}=2.83\times10^{-7}\text{ m}$ , the same answer both ways.

A wavelength of $283\text{ nm}$ is shorter than the visible range ( $\approx400$ – $700\text{ nm}$ ), so this photon lies in the ultraviolet region of the electromagnetic spectrum.

Part (d): Absorption line spectrum

When white light (a continuous range of wavelengths, and so of photon energies) passes through the gas, each ground-state atom can only absorb a photon whose energy exactly equals the difference between two of its allowed energy levels (for example the $4.40\text{ eV}$ found in part (b), corresponding to $283\text{ nm}$ ). Photons of that specific wavelength are absorbed, exciting electrons up to a higher level, while photons of other wavelengths pass straight through unaffected. The absorbed wavelengths are then missing (or greatly reduced in intensity) from the transmitted beam, appearing as dark absorption lines at those exact wavelengths in an otherwise continuous spectrum.

Final answers

(a) The atom’s energy can only take certain fixed, discrete values, not a continuous range.
(b) $\Delta E=\boxed{4.40}\text{ eV} = \boxed{7.04\times10^{-19}}\text{ J}$
(c) $f=\boxed{1.06\times10^{15}}\text{ Hz}$ , $\lambda=\boxed{283}\text{ nm}$ . Ultraviolet
(d) Only photons matching an exact energy-level gap are absorbed, removing those wavelengths and producing dark absorption lines.