find the work function of cesium and the stopping potential..?

Question

When cesium metal is illuminated with light of wavelength 300 nm, the photoelectrons emitted have a maximum kinetic energy of 2.23 eV find
the work function of cesium
the stopping potential if the incident of light has a wavelength of 400nm

Answer 1

• Photon frequency = c / λ = 3.0^8m/s / 300^-9 m = 1.0^15 Hz

• 2.23eV = 2.23*1.60^-19 = 3.57-19 J

Planck constant h = 6.625^-34 Js



From the equation hf = Φ + ½mv² (max) ..

Φ = hf - ½mv²



Φ = (6.625^-34 * 1.0^15) - 3.57^-19 .. .. ► Φ = 3.055-19 J .. (1.91eV)





2) Photon frequency = 3.0^8m/s / 400^-9m = 7.50^14 Hz



KE(max) = hf - Φ

KE = (6.625^-34 * 7.50^14 ) - 3.055^-19 = 1.91^-19 J



stopping voltage V [J/C] = 1.91^-19 J / 1.60^-19 C .. .. ► V = 1.20 V

Answer 2

Work Function Of Cesium

Answer 3

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RE:

find the work function of cesium and the stopping potential..?

When cesium metal is illuminated with light of wavelength 300 nm, the photoelectrons emitted have a maximum kinetic energy of 2.23 eV find

the work function of cesium

the stopping potential if the incident of light has a wavelength of 400nm

Answer 4

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wavelength= 2000A= 2000*10^(-10)m frequency=f=c/wavelength=1.5*10^15 stopping potential=4.21V hf=work function + maximumK.E of emitted electrns Energy of photon= 6.6*10^(-34)*1.5*10^15=9.9*10^(-19)Joule... K.E max= eV=1.67*10^(-19)*4.21joules=7.03*10^(-19... So, work function= hf - K.E max =9.9*10^(-19) - 7.03*10^(-19) = 2.87*10^(-19)J

Answer 5

Of course, you can assume the (slope) value = h/e=4.135708x10^ (-15) JsC^ (-1). Or you can calculate h/e = V1- V2 / (n1 - n2). That is why two values are given. However to calculate both work function and cut off frequency we are using the value of e. Therefore, it is enough to use one of the two sets and use the values of h and e.