(5%) Problem 6: In a photoelectric effect experiment you illuminate a metal with

Question

(5%) Problem 6: In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic energy of the photoelectrons to be 0.45 eV. Then you illuminate the same metal with light of a wavelength known to be 2/3 of the first wavelength and measure a maximum kinetic energy of 1.9 eV for the photoelectrons 50% Part (a) Find the first wavelength, in nanometers Grade Summary Deductions Potential 0% 100% Submissions Attempts remaining: Z (0% per attempt) detailed view tan() ( cosO cotan0 asin) acos0 456 atan)acotan) sinh0 cosh0 n cotanh0 Degrees O Radians 0 Submit Hint I give up! Hints: 2% deduction per hint. Hints remaining: 4 Feedback: 2% deduction per feedback 50% Part (b) Find the metal's work function, in electron volts

Explanation / Answer

A. Work function= hc/ lambda- k.e.

Wf=6.62×10^-34×3×10^8/lambda _ .45×1.6×10^-19

Again from 2nd wavelength

Wf = 6.62x10^-34×3×10^8×2/3lambda - 1.9×1.6×10^-19

Equating equation for both wavelength,

6.62x10^-34×3×10^8/3lambda = 2.6×1.6×10^-19

So, lambda = 6.62×10^-34×3×10^8/(3×2.6×1.6×10^-19)

=1.59x 10^-7= 159nm

B. Wf=

6.62×10^-34×3×10^8/(1.59×10^-7×1.6×10^-19) - .45

=7.35 ev

  

B.

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