Question

When you irradiate a metal with light of wavelength 441 nm in an investigation of the...

When you irradiate a metal with light of wavelength 441 nm in an investigation of the photoelectric effect, you discover that a potential difference of 1.07 V is needed to reduce the current to zero.

What is the energy of a photon of this light in electron volts?

energy of a photon:

Find the work function of the irradiated metal in electron volts.

work function:

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Answer #1

Here ,

wavelength = 441 nm

energy of photon , E = h * c/wavelength*e

E = 6.626 *10^-34 * 3 *10^8/(1.602 *10^-19 * 441 *10^-9)

E = 2.81 eV

the energy of photon is 2.81 eV

for the work function ,

work function = 2.81 - 1.07

work function = 1.74 eV

the work function is 1.74 eV

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