Answer:
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
