Students are studying the photoelectric effect using the apparatus shown in Figure 15. Figure 16 shows the results the students obtained for the maximum kinetic energy \((E_{\text{k max }})\) of the emitted photoelectrons versus the frequency of the incoming light. --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 2 WORK AREA LINES (style=lined) --- --- 0 WORK AREA LINES (style=blank) --- a. i. \(5.4 \times 10^{-15}\ \text{eV s}\) ii. \(811\ \text{nm}\) iii. \(1.9\ \text{eV}\). b. a.i. Planck’s constant \((h)\): → Equal to the gradient of the line when \(E_{\text{k max}}\) is graphed against frequency. \(\therefore h=\dfrac{\text{rise}}{\text{run}}=\dfrac{1.25-0}{6 \times 10^{14}-3.7\times 10^{-14}}=5.4 \times 10^{-15}\ \text{eV s}\) a.ii. → Max wavelength = minimum frequency of emitted photoelectron. \(\lambda=\dfrac{c}{f}=\dfrac{3 \times 10^8}{3.7 \times 10^{14}}=811\ \text{nm}\) a.iii. → The work function is the y-intercept of the graph, so by extending the graph as shown above, the work function is \(1.9\ \text{eV}\). b. → The new y-intercept for the graph will be \(-0.95\ \text{eV}\) → The gradient of the graph will remain the same (Planck’s constant)
On Figure 17, draw the line that would be obtained if a different metal, with a work function of \(\dfrac{1}{2} \phi\), were used in the photocell. The original graph is shown as a dashed line. (2 marks)
PHYSICS, M7 2023 HSC 9 MC
The graph shows the relationship between radiation intensity and wavelength for a black body at 4500 K.
Which statement describes the expected difference in the graph for a black body at 4000 K?
- Intensity at all wavelengths will be less.
- Intensity at all wavelengths will be greater.
- The peak intensity will occur at a higher frequency.
- The peak intensity will occur at a shorter wavelength.