An experimenter finds that no photo-electrons are emitted from tungsten unless the wavelength of light is less than 270nm. Her experiment will require photo-electrons of maximum kinetic energy 2.0eV. what frequency of light should be used to illuminate the tungsten?
To determine the frequency of light needed to illuminate the tungsten and generate photo-electrons with a maximum kinetic energy of 2.0eV, we can use the equation:
E = hf
where E is the energy of a photon, h is Planck's constant (6.626 x 10^-34 J·s), and f is the frequency of light.
First, we need to convert the maximum kinetic energy of photo-electrons from electron volts to joules:
1 eV = 1.6 x 10^-19 J
So, the maximum kinetic energy in joules is:
E = 2.0eV x 1.6 x 10^-19 J/eV
E = 3.2 x 10^-19 J
Now, we can rearrange the equation to solve for the frequency of light:
f = E / h
f = (3.2 x 10^-19 J) / (6.626 x 10^-34 J·s)
Calculating this, we find:
f = 4.838 x 10^14 Hz
Therefore, the frequency of light needed to illuminate the tungsten and generate photo-electrons with a maximum kinetic energy of 2.0eV is approximately 4.838 x 10^14 Hz.