The work function of a metal is ϕ eV. If λ nm UV light is incident on the photocathode, what is
the stopping voltage of the photoelectrons in volts?

The stopping voltage (\(V_{\text{stop}}\)) of the photoelectrons can be found using the **photoelectric equation**:

\[
E_{\text{kin}} = E_{\text{photon}} - \phi
\]

where:
- \(E_{\text{kin}}\) is the maximum kinetic energy of the photoelectrons,
- \(E_{\text{photon}}\) is the energy of the incident photon,
- \(\phi\) is the work function of the metal (in eV).

Since the stopping voltage corresponds to the kinetic energy of the electrons, we have:

\[
eV_{\text{stop}} = E_{\text{kin}}
\]

Thus, the stopping voltage is:

\[
V_{\text{stop}} = \frac{E_{\text{photon}} - \phi}{e}
\]

Now, let's break this down further:

### 1. Energy of the photon \(E_{\text{photon}}\)
The energy of the photon can be calculated using the wavelength of the incident UV light (\(\lambda\)) and Planck's equation:

\[
E_{\text{photon}} = \frac{hc}{\lambda}
\]

where:
- \(h = 6.626 \times 10^{-34} \, \text{J·s}\) is Planck's constant,
- \(c = 3.0 \times 10^8 \, \text{m/s}\) is the speed of light,
- \(\lambda\) is the wavelength in meters.

To convert this energy to electronvolts (eV), we use the relation \(1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J}\). Thus:

\[
E_{\text{photon}} = \frac{1240 \, \text{eV·nm}}{\lambda}
\]

### 2. Stopping voltage formula
Substituting \(E_{\text{photon}}\) into the formula for the stopping voltage:

\[
V_{\text{stop}} = \frac{\frac{1240}{\lambda} - \phi}{e}
\]

Since \(\phi\) is already in eV and \(e = 1 \, \text{eV}\), the equation simplifies to:

\[
V_{\text{stop}} = \frac{1240}{\lambda} - \phi
\]

### Conclusion
The stopping voltage \(V_{\text{stop}}\) in volts is given by:

\[
V_{\text{stop}} = \frac{1240}{\lambda} - \phi
\]

where:
- \(\lambda\) is the wavelength of the UV light in nm,
- \(\phi\) is the work function of the metal in eV.

Would you like to go into more detail, or clarify any part of the solution?

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Here are 5 related questions you might find interesting:
1. How does increasing the wavelength affect the stopping voltage?
2. What is the effect of increasing the work function \(\phi\) on the photoelectric effect?
3. Can the stopping voltage ever be negative? If yes, under what circumstances?
4. How is the photoelectric effect explained in quantum mechanics?
5. What happens if the photon energy is less than the work function?

**Tip:** The photoelectric effect provides direct evidence for the particle nature of light, confirming that light energy is quantized into photons.

The work function of a metal is ϕ eV. If λ nm UV light is incident on the photocathode, what is the stopping voltage of the photoelectrons in volts?

Learn how to calculate the stopping voltage of photoelectrons when UV light of wavelength λ nm is incident on a metal with a work function of ϕ eV. This problem covers the photoelectric effect, photon energy, and stopping voltage formula, suitable for undergraduate-level physics students.

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