Photoelectric Effect and Stopping Potential Calculation

The photoelectric effect describes the emission of electrons when light shines on a metal surface. The energy of the incident photon is used to overcome the work function of the metal (minimum energy required to eject an electron) and the remaining energy is given to the emitted electron as kinetic energy.

Einstein's photoelectric equation relates these quantities:

\begin{equation*} E = \phi + K_{max} \end{equation*}

where:

• $E$ is the energy of the incident photon.
• $\phi$ is the work function of the metal.
• $K_{max}$ is the maximum kinetic energy of the emitted photoelectron.

The maximum kinetic energy ($K_{max}$) is related to the stopping potential ($V_s$), which is the minimum negative potential applied to the collector plate to stop the most energetic electrons. This relationship is:

\begin{equation*} K_{max} = eV_s \end{equation*}

where $e$ is the elementary charge. Substituting this into the photoelectric equation, we get:

\begin{equation*} E = \phi + eV_s \end{equation*}

In this problem, both sodium and tungsten metals are illuminated by ultraviolet light of the same wavelength, meaning the energy of the incident photons ($E$) is the same for both metals.

Calculating Photon Energy using Tungsten Data

For tungsten, we are given:

• Work function ($\phi_W$) = 4.5 eV
• Stopping potential ($V_{sW}$) = 1.8 V

Using the equation $E = \phi + eV_s$, and noting that $eV_s$ has energy in Joules, while $V_s$ in Volts corresponds to energy $V_s$ in eV when multiplied by $e$, we can write the equation conveniently in terms of eV:

\begin{equation*} E = \phi_W + V_{sW} \quad (\text{when using } \phi \text{ in eV and } V_s \text{ in Volts/eV}) \end{equation*}

Plugging in the values for tungsten:

\begin{equation*} E = 4.5 \text{ eV} + 1.8 \text{ eV} \end{equation*}

\begin{equation*} E = 6.3 \text{ eV} \end{equation*}

So, the energy of the incident ultraviolet light photons is 6.3 eV.

Calculating Stopping Potential for Sodium

Now, we use the same photon energy ($E = 6.3$ eV) for sodium. For sodium, we are given:

• Work function ($\phi_{Na}$) = 2.3 eV
• Stopping potential for sodium ($V_{sNa}$) is what we need to find.

Using the photoelectric equation for sodium:

\begin{equation*} E = \phi_{Na} + eV_{sNa} \end{equation*}

Again, working in eV:

\begin{equation*} 6.3 \text{ eV} = 2.3 \text{ eV} + V_{sNa} \text{ eV} \end{equation*}

Rearranging the equation to solve for $V_{sNa}$:

\begin{equation*} V_{sNa} \text{ eV} = 6.3 \text{ eV} - 2.3 \text{ eV} \end{equation*}

\begin{equation*} V_{sNa} \text{ eV} = 4.0 \text{ eV} \end{equation*}

Thus, the stopping potential for sodium is 4.0 V.

Summary of Results

Using $E = \phi_{Na} + V_{sNa}$: $6.3 = 2.3 + V_{sNa}$, which gives $V_{sNa} = 6.3 - 2.3 = 4.0$ V.

The value of the stopping potential for sodium is 4.0 V.