An LED is a device that emits light of a particular frequency when the voltage applied across the LED reaches a certain value. This effect can be used to measure Planck’s constant.
In the circuit shown below the switch is closed and the potentiometer is adjusted until the LED just emits light. When this occurs the voltmeter reads 2.5 V. The uncertainty in the voltage reading is 0.2 V.
The LED operates in such a way that the electrons can give up the energy that they gain from the battery by emitting a photon. Assuming that all the energy gained by an electron is transferred to a photon and that the LED used in this experiment emits light of wavelength 480 nm calculate a value for Planck’s constant, together with an estimate of its uncertainty.
To calculate Planck's constant based on the information given, we can use the equation E = hc/λ, where E is the energy of a photon, h is Planck's constant, c is the speed of light, and λ is the wavelength of light.
First, let's find the energy of a photon using the formula E = qV, where q is the charge of an electron and V is the voltage across the LED. From the given information, the voltage across the LED is 2.5 V.
Next, we need to consider the uncertainty in the voltage reading, which is given as 0.2 V. To calculate the uncertainty in energy, we can use the formula δE = V × δq, where δE is the uncertainty in energy, V is the voltage, and δq is the uncertainty in charge.
Since the uncertainty in the voltage reading is 0.2 V, we can substitute these values into the formula to calculate the uncertainty in energy: δE = 2.5 V × 0.2 V = 0.5 V.
Now, we can calculate the energy of a photon: E = qV = (1.6 × 10^(-19) C) × (2.5 V) = 4 × 10^(-19) J.
To find Planck's constant, we can rearrange the equation E = hc/λ to h = Eλ/c. Using the values we have, h = (4 × 10^(-19) J) × (480 × 10^(-9) m) / (3 × 10^8 m/s) = 6.4 × 10^(-34) J·s.
Finally, to estimate the uncertainty in Planck's constant, we can use the formula δh = h × (δE/E + δλ/λ), where δh is the uncertainty in Planck's constant, δE is the uncertainty in energy, E is the energy of a photon, δλ is the uncertainty in wavelength, and λ is the wavelength of light.
Since the uncertainty in the voltage reading is 0.5 V, and the given wavelength is 480 nm with no uncertainty mentioned, we can substitute these values into the formula to calculate the uncertainty in Planck's constant.
δh = (6.4 × 10^(-34) J·s) × (0.5 V / 4 × 10^(-19) J + 0 / 480 × 10^(-9) m) = 8 × 10^(-36) J·s.
Therefore, the calculated value for Planck's constant is 6.4 × 10^(-34) J·s, with an estimated uncertainty of 8 × 10^(-36) J·s.