Question
(a) State an experimental phenomenon that provides evidence for:
(i) the particulate nature of electromagnetic radiation
……………………………………………………………………………………………………………………. [1]
(ii) the wave nature of matter.
……………………………………………………………………………………………………………………. [1]
(b) A particle of matter moves with momentum p.
(i) State the equation that gives the effective wavelength λ of the particle. State the name of any other symbols used. [2]
(ii) State the name given to the wavelength of the moving particle.
……………………………………………………………………………………………………………………. [1]
(c) Electrons are accelerated from rest through a potential difference (p.d.) of 4.8kV.
(i) Show that the final speed of the electrons is 4.1 × 107ms–1. [2]
(ii) Calculate the effective wavelength of a beam of electrons moving at the speed in (c)(i).
wavelength = …………………………………………….. m [2] [Total: 9]
Answer/Explanation
Ans
(a) (i) photoelectric effect
(a) (ii) electron diffraction
(b) (i) λ = h / p
h is the Planck constant
(b) (ii) de Broglie (wavelength)
(c )(i) ½mv2 = eV
½ × 9.11 × 10–31 × v2 = 1.60 × 10–19 × 4800 so v = 4.1 × 107 m s–1
(c) (ii) λ = h / mv = 6.63 × 10–34 / (9.11 × 10–31 × 4.1 × 107)
= 1.8 × 10–11
Question
(a) Electromagnetic radiation of a single constant frequency is incident on a metal surface. This causes an electron to be emitted.
Explain why the maximum kinetic energy of the electron is independent of the intensity of the incident radiation.
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………………
…………………………………………………………………………………………………………………………… [3]
(b) Ultraviolet radiation of wavelength 250nm is incident on the surface of a sheet of zinc. The maximum kinetic energy of the emitted electrons is 1.4eV.
Determine, in eV:
(i) the energy of a photon of the ultraviolet radiation
energy = ……………………………………………. eV [3]
(ii) the work function energy of the surface of the zinc.
energy = …………………………………………….
Answer/Explanation
Ans:
(a)
- frequency determines energy of photon
- intensity determines number of photons (per unit time)
- intensity does not determine energy of a photon
Any two points, 1 mark each
kinetic energy (of the electron) depends on the energy of one photon
(b)(i) \(E = hc / λ\)
or
\(E = hf\) and \(c = fλ\)
\(E = (6.63× 10^{–34}× 3.00× 10^8) / (250× 10^{–9})\)
\((= 7.96× 10^{–19} J)= 5.0 eV\)
(b)(ii) \(E_{MAX}\) = photon energy – work function