An LED that emits light of wavelength λ = 667 nm illuminates the phototube. A reverse bias is applied to the phototube (a voltage that opposes the current flow). This is adjusted carefully until the photocurrent drops to zero. The stopping potential is found to be Vo = 0.43 V.The LED can now be changed and the process repeated for a different λ. Then the data can be plotted to determine Planck's constant, h.For the data collected above what value would you plot on the x-axis of your graph? (Use c = 3.00 x 108m/s.)Express your answer in SI units to 3 significant figures.Entering numbers in scientific notation: Example: 1.45 x 10-9 should be entered as 1.45E-9

The slope of the stopping voltage versus frequency graph was found to be m = 3.78 x 10-15 Vs. Use this value to determine Planck's constant.Express you answer to 3 significant figures in SI units. (Use a value of e = 1.6 x 10-19C for the charge on an electron.)Entering numbers in scientific notation: Example: 1.45 x 10-9 should be entered as 1.45E-9

The slope of the stopping voltage versus frequency graph was found to be m = (4.36 ± 0.75) x 10-15 Vs. Using this value to determine Planck's constant we obtain h = 6.976 x 10-34Js.Determine the uncertainty in h. Express you answer to 2 significant figures in SI units.Entering numbers in scientific notation: Example: 1.45 x 10-9 should be entered as 1.45E-9

In the experiment performed to calculate the value of Plank's constant, the distance of the light source from a photo diode was varied and the corresponding photo current was recorded. Values of current and distance were used in derived formula to calculate planks constant.Select one:TrueFalse

Monochromatic radiation emitted when electron on hydrogen atom jumps from first exited to the grounds state irradiates a photosnesitive material. The stopping potential is measured to be 3.57V. The threshold frequency of the material is :A1.6×1015HzCorrect AnswerB2.5×1015HzYour AnswerC4×1015HzD5×1015Hz

The equipment is set up as in the figure below. Light from the green LED illuminates the phototube. The applied voltage dial is adjusted and measurements of V1 and V2 are made:V1 = 4.41 VV2 = 0.51 VWhat is the photocurrent in the circuit? Express your answer in SI units to 3 significant figures.Entering small numbers: Example: 1.45 x 10-6 should be entered as 1.45E-6