In a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic energy of the photoelectrons to be 0.75 eV. Then you illuminate the same metal with light of a wavelength known to be 2/3 of the first wavelength and measure a maximum kinetic energy of 1.8 eV for the photoelectrons. Required:
a. Find the first wavelength, in nanometers
b. Find the metals work function, in electron volts

A.) 1658 nm

B.) 1.05 ev

Explanation:

Given that in a photoelectric effect experiment you illuminate a metal with light of an unknown wavelength and measure the maximum kinetic energy of the photoelectrons to be 0.75 eV. Then you illuminate the same metal with light of a wavelength known to be 2/3 of the first wavelength and measure a maximum kinetic energy of 1.8 eV for the photoelectrons.Required:

a. Find the first wavelength, in nanometers

The formula to use is:

E = hf

Where f = c/ wavelength.

C = speed of light.

Convert the electron volts to Joule.

0.75 × 1.6 × 10^-19 = 1.2 × 10^-19 J

Substitute the values into the formula

1.2 × 10^-19 = (6.63 × 10^-34 × 3 × 10^8) ÷ wavelength

Wavelength = (6.63×10^-34 × 3×10^8) ÷ 1.2×10^-19

Wavelength = 1.6575 × 10^-6 m

Therefore, the wavelength of the first light in nanometers will be

Wavelength = 1658 nm

b. Find the metals work function, in electron volts

The metal works function will be:

WF = 1.8 - 0.75 = 1.05 eV

photocells convert light energy to electricity