Hi i could use some help to fill out the green section of this
lab I did I think I provided everything needed thanks
alot.
Hi I Could Use Some Help To Fill Out The Green Section Of This Lab I Did I Think I Provided Everything Needed Thanks Alo 1 (127.3 KiB) Viewed 27 times
Mercury Vapor Lamp 578 Order Kind of Light Wavelength Filter First Yellow Yellow First Yellow 578 Yellow First Yellow 578 Yellow First Yellow 578 Yellow First Yellow 578 Yellow First Green 546.074 Green First Green 546.074 Green First Green 546.074 Green First Green 546.074 Green First Green 546.074 Green First Blue 435.074 None First Violet 404.656 None First Ultraviolet 365.483 None Second Yellow 578 None Second Green 546.074 None Second Blue 435.835 None Second Violet 404.656 None Second Ultraviolet 365.483 None Transmission Filter Stopping Potential Energy v Intensity 100 0.73 80 0.73 60 0.72 Do you find such a relationship? 40 0.72 20 0.72 100 0.76 80 0.75 60 0.74 Do you find such a relationship? 40 0.73 20 0.72 100 1.26 100 1.59 100 1.86 100 1.62 100 1.63 100 1.24 100 1.42 100 1.86 Plot the stopping potential vs frequency. Frequence=x sloper Plants Constant Slope of the line h - Experimental h - Accepted Compare the h's O Work Function
E - = E = hf = h. A = (1) where h is Planck's constant, h = 6.63 x 10-34 [J s]. =
The energy of the photoelectrons comes from a photon of light so conservation of energy for the photoelectrons looks like E = hf = = 1 my2 + 0 2 (2) In your h/e apparatus, the photoelectrons are released from the cathode and will flow towards the anode (positive terminal) as a photocurrent. You will measure the kinetic energy of the photoelectrons from Equation (2) by finding the voltage necessary to stop the photoelectrons. The work needed to stop the photoelectrons is 1 2 Work = AK = 0 = =qV =-eV = (3) or 1 my2 = eV žm (4) so the voltage that stops the photoelectric current tells you the kinetic energy of the photoelectrons and Equation (2) becomes hf = eV + 0 = (5 ) and solving for the Voltage V, we have
Φ v-C) - If e е (6)
Join a community of subject matter experts. Register for FREE to view solutions, replies, and use search function. Request answer by replying!