The work function of a material is 3.5 eV. If the material is the illuminated with monochromatic light (l = 300 nm), what are:
a) the stopping potential
b) the cutoff frequency
google "Einstein, photoelectric effect"
To find the stopping potential and cutoff frequency, we need to use the relationship between the work function, the frequency of the incident light, and the energy of a photon.
a) The stopping potential is the minimum potential difference that must be applied across a metal surface to completely stop the emitted electrons. We can calculate it using the formula:
Stopping potential (V) = (Planck's constant x Frequency) / Electronic charge
First, let's calculate the energy of a photon using the wavelength of monochromatic light given (l = 300 nm). The energy (E) of a photon can be calculated using the equation:
Energy (E) = (Planck's constant x Speed of light) / Wavelength
Planck's constant (h) = 6.63 x 10^-34 J s
Speed of light (c) = 3 x 10^8 m/s
Convert the wavelength to meters: l = 300 nm = 300 x 10^-9 m
E = (6.63 x 10^-34 J s x 3 x 10^8 m/s) / (300 x 10^-9 m)
E ≈ 6.63 x 10^-19 J
Next, we can calculate the frequency (f) of the incident light using the equation:
Energy (E) = Planck's constant x Frequency
Rearranging the equation, we get:
Frequency (f) = Energy (E) / Planck's constant
f ≈ (6.63 x 10^-19 J) / (6.63 x 10^-34 J s)
f ≈ 10^15 Hz
Finally, we can calculate the stopping potential:
Stopping potential (V) = (Planck's constant x Frequency) / Electronic charge
V ≈ (6.63 x 10^-34 J s x 10^15 Hz) / 1.6 x 10^-19 C
V ≈ 41.4 V
Therefore, the stopping potential is approximately 41.4 volts.
b) The cutoff frequency (f_cutoff) is the minimum frequency of light required to emit electrons from a material. To calculate it, we can rearrange the equation above:
Frequency (f_cutoff) = (Work function) / Planck's constant
Given that the work function is 3.5 eV, we need to convert it to joules using the conversion factor:
1 eV = 1.6 x 10^-19 J
Work function (W) = 3.5 eV x 1.6 x 10^-19 J/eV
W ≈ 5.6 x 10^-19 J
Now we can calculate the cutoff frequency:
Frequency (f_cutoff) = (Work function) / Planck's constant
f_cutoff ≈ (5.6 x 10^-19 J) / (6.63 x 10^-34 J s)
f_cutoff ≈ 8.45 x 10^14 Hz
Therefore, the cutoff frequency is approximately 8.45 x 10^14 Hz.