(a) Briefly explain the following terms: (i) emission line spectra; (ii) line absorption spectra. (b) Draw a labeled diagram showing the structure of a simp...

Answer Details

(a)(i) Emission line spectrum

An emission line spectrum consists of separate bright lines on a dark background. Each line has a definite wavelength. It is produced when excited atoms emit photons as electrons fall from higher to lower energy levels.

(a)(ii) Line absorption spectrum

A line absorption spectrum consists of dark lines in an otherwise continuous spectrum. It is produced when light passes through a cooler gas: atoms in the gas absorb photons of particular wavelengths to raise electrons to higher energy levels.

(b) Simple photoemissive photocell

Light must have a frequency at least equal to the threshold frequency of the photosensitive cathode. The light then causes electrons to be emitted from the cathode by the photoelectric effect. Since the anode is positive relative to the cathode, it attracts and collects these electrons. Their movement through the external circuit produces a photocurrent, detected by the galvanometer. Increasing light intensity increases the number of emitted electrons per second, so the photocurrent increases.

(c)(i) Evidence that X-rays are waves

• X-rays undergo diffraction, for example when diffracted by the regularly spaced planes of atoms in a crystal.
• X-rays produce interference patterns. Diffraction and interference are characteristic wave behaviours.

(c)(ii) Uses of X-rays other than in medicine

• Industrial radiography: detecting internal cracks or flaws in welded joints and metal castings.
• Determining crystal structure by X-ray diffraction.

(d) Energy change and wavelength

There is an inconsistency in the wording: an electron moving from \( -1.6\ \text{eV} \) to \( -1.4\ \text{eV} \) moves to a higher energy level, because \( -1.4\ \text{eV} \) is greater than \( -1.6\ \text{eV} \). Therefore, the electron must absorb radiation; it cannot emit radiation during this transition.

The magnitude of the energy absorbed is:

\[ \Delta E=(-1.4)-(-1.6)=0.2\ \text{eV} \] \[ \Delta E=0.2(1.6\times10^{-19})=3.2\times10^{-20}\ \text{J} \]

Using \(E=\dfrac{hc}{\lambda}\):

\[ \lambda=\frac{hc}{E} =\frac{(6.6\times10^{-34})(3.00\times10^8)}{3.2\times10^{-20}} \] \[ \lambda=6.19\times10^{-6}\ \text{m}\approx6.2\times10^{-6}\ \text{m} \]

Thus, for the levels printed in the question, radiation of energy \(0.20\ \text{eV}\), or \(3.2\times10^{-20}\ \text{J}\), and wavelength \(6.2\times10^{-6}\ \text{m}\) is absorbed. Emission would occur only for the reverse transition, from \( -1.4\ \text{eV} \) to \( -1.6\ \text{eV} \).