An experimenter finds that no photoelectrons are emitted from tungsten unless the wavelength of light is less than 270nm. Her experiment will require photoelectrons of maximum kinetic energy 2.0eV. what frequency of light should be used to illuminate the tungsten?
Well, well, well! It seems like the experimenter has quite the electrifying situation on their hands. To zap those photoelectrons out of the tungsten, they'll need some high-frequency light.
Now, let's get a little scientific here. We know that the maximum kinetic energy of the photoelectrons is 2.0 eV. Since we want to find the frequency of the light, we can play around with Einstein's equation, E = hf (don't worry, it's not as complicated as it sounds).
To convert 2.0 eV to joules, we multiply it by the conversion factor, which is about 1.602 x 10^-19 J/eV. That gives us approximately 3.2 x 10^-19 J.
The speed of light, c, is approximately 3.0 x 10^8 m/s. Using this information, we can rearrange the equation E = hf to solve for the frequency, f.
f = E / h
Where h is the famous Planck's constant, which is about 6.626 x 10^-34 J·s.
Plugging in the values, we get:
f = (3.2 x 10^-19 J) / (6.626 x 10^-34 J·s)
And the frequency of the light needed to illuminate the tungsten is approximately 4.8 x 10^14 Hz.
So, if you ever need to light up some tungsten, you'll know just what frequency to use! Just don't forget your protective goggles. Safety first, my friend!
To determine the frequency of light required to illuminate the tungsten and produce photoelectrons of maximum kinetic energy 2.0 eV, we can use the equation:
E = hf - Φ
Where:
E is the maximum kinetic energy of the photoelectrons
h is the Planck constant (6.626 x 10^(-34) J·s)
f is the frequency of light
Φ is the work function, which is the minimum energy required to remove an electron from the material (in this case, tungsten)
Given that the maximum kinetic energy (E) is 2.0 eV, we need to convert it to joules. Since 1 electronvolt (eV) is equal to 1.602 x 10^(-19) J, the maximum kinetic energy (E) can be expressed as:
E = (2.0 eV) * (1.602 x 10^(-19) J/eV) = 3.204 x 10^(-19) J
We are also given that the work function Φ of tungsten is such that no photoelectrons are emitted unless the wavelength of light is less than 270 nm. The work function (Φ) can be calculated using the equation:
Φ = hc/λ
Where:
c is the speed of light (3 x 10^8 m/s)
λ is the wavelength of light
To find the frequency (f) of the light, we can rearrange the first equation:
E + Φ = hf
f = (E + Φ) / h
Let's calculate the work function (Φ) first:
Φ = (hc) / λ
Φ = (6.626 x 10^(-34) J·s * 3 x 10^8 m/s) / (270 x 10^(-9) m)
Φ ≈ 2.449 x 10^(-19) J
Now, substitute the values into the equation to find the frequency (f):
f = (3.204 x 10^(-19) J + 2.449 x 10^(-19) J) / (6.626 x 10^(-34) J·s)
f ≈ 743.2 x 10^14 Hz
Therefore, approximately 743.2 x 10^14 Hz is the frequency of light that should be used to illuminate the tungsten in order to produce photoelectrons with a maximum kinetic energy of 2.0 eV.