%IMG% The diagram above illustrates a structure of a typical photocell. (i) Identify each of the parts labelled A and B. (ii) State one function each of A a...

Assessment: WAEC SSCE - Physics - 2019 (Essay) Subject: Physics

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%IMG%

The diagram above illustrates a structure of a typical photocell.

(i) Identify each of the parts labelled A and B.

(ii) State one function each of A and B.

(iii) Einstein’s photoelectric equation can be written as \(E = hf - W_o\). State what each of the terms \(E\), \(hf\) and \(W_o\) represent.

(b) A photon is incident on a metal whose work function is \(1.32\ \text{eV}\). An electron is emitted from the surface with a maximum kinetic energy of \(1.97\ \text{eV}\). Calculate the frequency of the photon. \([1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}]\)

(c)(i) Define half-life of a radioactive element.

(ii) Sketch a graph of the relation \(N = N_0e^{-\lambda t}\) and indicate the half-life.

Answer Details

The photocell has the structure shown below.

figure
Structure of a typical vacuum photocell: A = emitter (photocathode), B = collector (anode), with the external microammeter circuit.

(a)(i) Identification of the labelled parts

  • A - the emitter (photocathode): the large curved metal plate coated with a photosensitive material.
  • B - the collector (anode): the small metal rod placed in front of the cathode.

(a)(ii) Function of each part

  • A emits electrons when light of sufficient frequency falls on it.
  • B attracts and collects the emitted electrons, so completing the circuit and allowing a photocurrent to flow.

(a)(iii) Meaning of the terms in \(E = hf - W_0\)

  • \(E\) - the maximum kinetic energy of the emitted photoelectrons.
  • \(hf\) - the energy of the incident photon (\(h\) = Planck's constant, \(f\) = frequency).
  • \(W_0\) - the work function of the metal, the minimum energy needed to release an electron from its surface.

(b) Frequency of the photon

The photon energy equals the maximum kinetic energy of the electron plus the work function:

\[ hf = KE_{max} + W_0 = 1.97 + 1.32 = 3.29\ \text{eV} \]

Converting to joules:

\[ hf = 3.29 \times 1.6 \times 10^{-19} = 5.264 \times 10^{-19}\ \text{J} \]

Therefore the frequency is:

\[ f = \frac{hf}{h} = \frac{5.264 \times 10^{-19}}{6.6 \times 10^{-34}} = 7.98 \times 10^{14}\ \text{Hz} \]

(c)(i) Definition of half-life

The half-life of a radioactive element is the time taken for half the nuclei (atoms) originally present in a sample to decay.

(c)(ii) Graph of \(N = N_0 e^{-\lambda t}\)

Taking an initial number \(N_0 = 800\) undecayed nuclei, the following readings of \(N\) against time \(t\) are obtained:

Time \(t\) (s) 0 1 2 3 4 5 6 7 8
Undecayed nuclei \(N\) 800 566 400 283 200 141 100 71 50

Plotting these readings gives the exponential decay curve:

graph
Exponential decay curve. N falls from N0 = 800 to N0/2 = 400 at t = 2 s, giving a half-life of 2 s (indicated by the drop from 400 to the time axis).

The half-life is read where the number of undecayed nuclei falls to \(N_0/2 = 400\). From the table and graph this occurs at \(t = 2\ \text{s}\); the count halves again to 200 at \(t = 4\ \text{s}\) and to 100 at \(t = 6\ \text{s}\), confirming a constant half-life:

\[ t_{1/2} = 2\ \text{s} \]

This value is indicated on the graph by the dashed lines drawn from \(N = 400\) across to the curve and down to the time axis.

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