What is the threshold frequency for the photoelectric effect on lithium (phi= 2.93eV)? What is the stopping potential if the wavelength of the incident light is 400nm?
To find the threshold frequency for the photoelectric effect on lithium, you can use the equation:
E = hf
Where:
E is the energy of a photon (in eV)
h is the Planck's constant (approximately 4.136 × 10^-15 eV s)
f is the frequency of the light (in Hz)
Since the energy of a photon is given by E = phi (work function), we can rewrite the equation:
phi = hf
To find the threshold frequency for lithium, we can rearrange the equation:
f = phi/h
Substituting the given values:
phi = 2.93 eV
h = 4.136 × 10^-15 eV s
f = 2.93 eV / 4.136 × 10^-15 eV s
Calculating this will give you the threshold frequency for the photoelectric effect on lithium.
To find the stopping potential if the wavelength of the incident light is 400 nm, we can use the equation:
lambda = c / f
Where:
lambda is the wavelength of the light (in meters)
c is the speed of light (approximately 3 × 10^8 m/s)
f is the frequency of the light (in Hz)
Rearranging the equation, we get:
f = c / lambda
Substituting the given values:
lambda = 400 nm = 400 × 10^-9 m
c = 3 × 10^8 m/s
f = 3 × 10^8 m/s / (400 × 10^-9 m)
Calculating this will give you the frequency of the light.
To find the stopping potential, we can use the equation:
V = (hf - phi) / e
Where:
V is the stopping potential (in volts)
h is the Planck's constant (approximately 4.136 × 10^-15 eV s)
f is the frequency of the light (in Hz)
phi is the work function (in eV)
e is the elementary charge (approximately 1.602 × 10^-19 C)
Substituting the given values:
h = 4.136 × 10^-15 eV s
f (calculated) = ...
phi = 2.93 eV
e = 1.602 × 10^-19 C
V = (f (calculated) × h - phi) / e
Calculating this will give you the stopping potential.
To determine the threshold frequency for the photoelectric effect on lithium, we can use the equation:
\(E_{\text{photon}} = \phi + \frac{hc}{\lambda}\)
Where:
\(E_{\text{photon}}\) is the energy of the incident photon,
\(\phi\) is the work function of the material (in this case, lithium),
\(h\) is Planck's constant (\(6.626 \times 10^{-34}\, \text{J s}\)),
\(c\) is the speed of light (\(3 \times 10^8\, \text{m/s}\)),
\(\lambda\) is the wavelength of the incident light.
Given that \(\phi = 2.93\,\text{eV}\), we need to convert it to joules by using the conversion factor: \(1\, \text{eV} = 1.6 \times 10^{-19}\, \text{J}\).
So \(\phi = 2.93 \times 1.6 \times 10^{-19}\, \text{J}\).
Now, we can substitute the values into the equation:
\(E_{\text{photon}} = 2.93 \times 1.6 \times 10^{-19} + \frac{(6.626 \times 10^{-34})(3 \times 10^8)}{(400 \times 10^{-9})}\)
After performing the calculation, we can find the energy of the photon.
To determine the stopping potential, we can use the equation:
\(V_{\text{stop}} = \frac{E_{\text{photon}}}{e}\)
Where:
\(V_{\text{stop}}\) is the stopping potential,
\(E_{\text{photon}}\) is the energy of the incident photon calculated earlier,
\(e\) is the elementary charge (\(1.6 \times 10^{-19}\,\text{C}\)).
Now, we can substitute the values into the equation and calculate the stopping potential.