A light-emitting diode (LED) emits light over a narrow range of wavelengths - AQA - A-Level Physics - Question 2 - 2021 - Paper 3

One possible disadvantage of using the fifth-order maximum is that the $5^{th}$ maximum may be less visible or fainter compared to lower-order maxima. This could lead to less accurate measurements due to lower intensity and visibility issues.

To find $V_A$ for $ext{L}_R$, we need to extrapolate the linear region of the current-voltage characteristic for $ext{L}_R$ from Figure 4. The intersection point where this extrapolation meets the horizontal axis indicates the activation voltage. Based on the graph, we observe that this point occurs at approximately $V_A = 1.80 ext{ V}$.

From the equation that describes the activation voltage, we use the known value of $V_A$ for $ext{L}_G$ at 2.00 V to calculate the Planck constant $h$: Using the formula:
V_A = rac{hc}{e ext{ } ext{ }}
We can rearrange this to solve for $h$: h = rac{V_A imes e}{c}
Substituting values with:

• $V_A = 2.00 ext{ V}$,
• $c = 3.00 imes 10^8 ext{ m/s}$,
• $e = 1.60 imes 10^{-19} ext{ C}$
  We find: h = rac{(2.00)(1.60 imes 10^{-19})}{(3.00 imes 10^8)} = 1.06 imes 10^{-34} ext{ Js}.
  Thus, the Planck constant can be approximated as $h ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } ext{ } 1.06 imes 10^{-34} ext{ J s}$.

To deduce the minimum value of the resistor $R$, we start from Ohm's law, acknowledging that: $V = IR$
From the circuit information, we have:

• The total emf is $6.10 ext{ V}$,
• The maximum current through the LED $ext{L}_R$ is $21.0 ext{ mA} = 0.021 ext{ A}$.
  Rearranging Ohm's law gives us: R = rac{V}{I}
  Thus, substituting in the values: