The apparatus shown above is used in determining the work function of a particul

Question

The apparatus shown above is used in determining the work function of a particular metal using the photoelectric effect. The experiment is set up with an ammeter A and a variable power supply. A light source that emits photons of 400 nm is used. The emf epsilon provided by the power supply is slowly increased from zero until the ammeter shows that no current is flowing between the plates The magnitude of the emf is 0.65 V when the current stops flowing. Determine the frequency of the incident photons Calculate the work function of the metal Calculate the frequency of light at or below which no current will flow If the intensity of the incident light is increased and the wavelength stays the same, does the magnitude of the emf needed to stop the current increase, decrease, or remain the same? Justify your response. If the wavelength of the light is decreased while the intensity of the incident beam remains fixed, does the number of photoelectrons emitted from the metal surface per unit time increase, decrease, or remain the same? Justify your response.

Explanation / Answer

a) Frequency is c/wavelength = 3*10^8/(400*10^-9) = 7.5*10^14 Hz

b) Energy incident, E = h*frequency = 6.6*10^-34* 7.5*10^14 /1.6*10^-19 eV = 309.37 eV

Work function of metal =E-Stopping Potential= 309.37 eV - 0.65 = 308.725 eV

c) frequency at or below which no current will flow is 308.725*1.6*10^-19/(6.6*10^-34) = 74.84*10^15 Hz

d) No, It will remain the same as wavelength remains constant

e) It will remain the same as number of photoelectrons depends on intensity

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