Question

function f(x)? Determine an expression for AB- BA in terms of f(x). 5. Consider the photoelectric effect for cesium metal. Experimental data for the kinetic energy of ejected electrons as a function of the wavelength of the incident radiation are shown below: 400 1.004 500 0.383 100 200 300 2 (nm) KE eV) 030 4.108 2.038 By constructing a linear plot using a graphing program or graph paper, calculate the value for h (Planck's constant in J.s) and the work function for cesium (in eV). Include a copy of your linear plot, and note that l eV=1.6x10-19J.

Answer 1

please ignore the extra rough line in th graph

A^= n B^= partial differential/partial differential n rightdoublearrow [A^B^- B^A^] f(n) rightdoublearrow[n partial differential/partial differential n - partial differential/partial differential n n] f(n) {f' (n) = partial differential/partial differential n f(n) rightdoublearrow n partial differential/partial differential n f(n) - partial differential/partial differential n n f(n) rightdoublearrow n f' (n) - [n P'(n) + f(n) middot 1] rightdoublearrow f(n) work function of caesium = ? In photoelectric effect when a radiation fall on any surface the with energy hv after some absorption due to work function the rest is converted into kinetic energy of e^- h v = w + K.E v = c/lambda h c/lambda = w + K.E K.E = h c/lambda - w because y = mr + c \r\n Now slope of the graph = 6.5561 times 10^-25 - 9 = 6.5561 times 10^-34 and intercept = 2 e v Actual value of h = 6.626 times 10^-34 our value is very near the actual volume Actually it is having = Vertical-bar 6.6261 = 6.5561 Vertical-bar /6.6262 times 100 = 1.06 % error only