REZA KHATAMI - Activity--Measuring Planck's Constant

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University of Toronto *

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125

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Chemistry

Date

Feb 20, 2024

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docx

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SPH4U Name:___________ Activity–Measuring Planck’s Constant Key Equations Materials ● Set of 5 LEDs of different, known frequencies ● 4 AA batteries ● voltmeter ● Potentiometer with 330 Ω resistor Procedure CAUTION: Do not stare directly at a brightly lit LED. 1. Turn the potentiometer knob all the way clockwise. This corresponds to 0 V. 2. Insert an LED into the device. 3. Slowly increase the potential difference across the LED by turning the potentiometer knob counterclockwise until the LED just begins to glow. It is recommended to darken the room when attempting to measure the LED brightness. 4. If the LED does not light up, reverse the LED legs. Recall that the long leg is the positive (anode). 5. Record the potential difference at which this happens by measuring across the terminals. Go backwards and forwards past the point at which the LED just begins to glow a few times to locate it as accurately as possible. 6. Repeat Step 4 for all the other LEDs. Always turn the potentiometer knob fully clockwise before changing LEDs so the initial voltage across each LED is 0 V. 7. Summarize your results in a table similar to the table below: Colour of LED Red Amber Yellow Green Blue Frequency (× 10 14 Hz ) 4.54 5.00 5.08 5.31 6.38 Potential difference (V) 1.35 1.43 1.52 1.55 2.15

SPH4U Name:___________ Analysis 1. Plot a graph of potential difference ( y -axis) versus frequency ( x -axis). 2. Insert the line of best fit and determine its slope. Calculate Planck’s constant using the relationship: ΔV =( h e ) f , so slope = h e slope ( e ) = h 4.54 x 10 -15 (1.602 x 10 -19 J) = h h = 7.27 x 10 -34 Hint: If your number is way off, check your slope. Remember that your frequency is measured in 10 14 Hz. Evaluation The accepted value for Planck’s constant, h , is 6.63 × 10 –34 J ⋅ s. What is the percentage error in your calculated value for Planck’s constant? What might have caused this error? 1. Outside light sources, such as those from phone screens and flashlights, could have attributed to the light

SPH4U Name:___________ being given more volts since it was flushed out by said lights. 2. The LED could have been lit at fewer volts but as it was imperceptible to the naked eye, we had to increase the voltage, making the percentage error higher as it wasnt the exactly the minimum number of volts required to light the LED. % error = ¿ 6.63 x 10 − 34 −( 7.27 x 10 − 34 )∨ ¿ 6.63 x 10 − 34 x 100% ¿ = 9.65% Questions 1. A green laser pointer produces 530 nm light with a power rating of 1.0 mW. How many photons does the laser produce each second? (Recall that a watt is a joule per second.) c = 3.00 x 10 8 m/s h = 6.63 x10 -34 Js = 5.30 x 10 𝝀 -7 m 1.0 mW = 1.0 x 10 -3 W → 1.0 x 10 -3 J/s f = c / 𝝀 f = (3.00 x 10 8 m/s) / (5.30 x 10 -7 m) f = 5.66 x 10 14 Hz E = hf E = (6.63 x10 -34 Js) x (5.66 x 10 14 Hz) E = 3.75 x 10 -19 J Number of photons per second: = (1.0 x 10 -3 J/s) / (3.75 x 10 -19 J) = 2.667 x 10 15 ∴ The laser produced 2.7 x 10 15 photons per second 2. Weather reports monitor ultraviolet (UV) light levels. Why is UV light a concern? Why is UV light beneficial for humans? Concerns: a. Can damage skin and eyes: Can cause, sun burns, skin damage (Wrinkles, spots), skin cancer and temporary or permanent blindness b. UV light can damage DNA and suppress your immune system c. UV light slows down the rate of photosynthesis in aquatic ecosystems. Uses: a. Stimulates vitamin D production b. Kills bacteria, fungus, microbes in the air i. Used in HVAC systems for that reason ii. UV lights become self cleaning

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