Cesium metal (work function, = 2.1 eV) is to be used as the photocathode materia

Question

Cesium metal (work function, = 2.1 eV) is to be used as the photocathode material in a photo-emissive electron tube because electrons are relatively easily removed from a cesium surface.

1. What is the longest wavelength of radiation, which can result in photoemission (provide your answer in nm)?

2. If ultraviolet radiation of wavelength 300 nm is incident onto the Cs photocathode, what will be the kinetic energy of the photo-emitted electrons in eV? (provide your answer in eV)

3. Consider using one of the following materials as photocathode. If the 300 nm ultraviolet radiation is used, for which material there will be NO photoelectric current? (the element's name - i.e. Magnesium etc.)

Element       Work Function(eV)

Calcium       2.9

Magnesium 3.68

Potassium   2.3

Iron             4.5

Explanation / Answer

given

workfunction, Wo = 2.1 eV

1) for lonegst wavelength, E_photon = Wo

h*c/lamda_max = Wo

lamda_max = h*c/Wo

= 6.626*10^-34*3*10^8/(2.1*1.6*10^-19)

= 5.91*10^-7 m

= 591 nm

2) use, E_photon = Wo + KEmax

==> KEmax = E_photon - Wo

= h*c/lamda - Wo

= 6.626*10^-34*3*10^8/(300*10^-9) - 2.1 eV

= 6.626*10^-19 J - 2.1 eV

= 6.626*10^-19/(1.6*10^-19) eV - 2.1 eV

= 4.14 eV - 2.1 eV

= 2.04 eV

3) Workfunction of the material, Wo = h*c/lamda

= 6.626*10^-34*3*10^8/(300*10^-19)

= 4.14 eV

In the given data Iron has Wo = 4.5 eV > 4.14 eV

so, Iron will be the material

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