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 find the frequency of light needed to illuminate the tungsten, we can use the relationship between frequency and wavelength of light. The formula is:
c = λν
where:
- c is the speed of light in a vacuum (approximately 3.00 x 10^8 m/s),
- λ is the wavelength of light (in meters), and
- ν is the frequency of light (in Hz).
First, we need to convert the maximum kinetic energy from electron volts (eV) to joules (J). We know that 1 eV is equal to 1.6 x 10^-19 J.
So, the maximum kinetic energy in joules, E_max, is given by:
E_max = 2.0 eV x 1.6 x 10^-19 J/eV.
Now, we can use the relationship between energy and frequency to connect the maximum kinetic energy to the frequency of light:
E_max = hν
where:
- h is Planck's constant (approximately 6.63 x 10^-34 J·s), and
- ν is the frequency of light (in Hz).
Now, we can solve for the frequency of light, ν:
ν = E_max / h.
Substituting the values, we have:
ν = (2.0 eV x 1.6 x 10^-19 J/eV) / (6.63 x 10^-34 J·s).
By performing this calculation, the result will give us the frequency of light required to illuminate the tungsten and cause the emission of photo-electrons.