The stopping potential for electrons emitted from a photosensitive surface illuminated by light of w

Q: Why do we use $1240 \text{ eV}\cdot\text{nm}$ for $hc$?

Using $hc \approx 1240 \text{ eV}\cdot\text{nm}$ is a standard, highly efficient shortcut in modern physics. It allows you to directly plug in the wavelength in nanometers and get the energy in electron-volts (eV), saving you from tedious and error-prone conversions involving Planck's constant and the speed of light in standard SI units.

Q: Why did we subtract the two equations instead of finding the work function first?

The work function $\phi_0$ is a property of the metal surface, which remains the same in both cases. While you *could* calculate it first, subtracting the equations elegantly eliminates this unknown variable in one step, reducing the chance of rounding errors and saving precious time during an exam.

Q: Does the stopping potential depend on the intensity of the incident light?

No, the stopping potential depends strictly on the frequency (or wavelength) of the incident light and the nature of the material. Increasing the intensity increases the number of emitted electrons (the photocurrent) but does not change their maximum kinetic energy.

Q: What if the calculated wavelength was longer than the threshold wavelength?

If the incident wavelength is longer than the threshold wavelength, the energy of the photons would be less than the work function. In that scenario, no photoelectric emission would occur at all, and the concept of a stopping potential wouldn't even apply.

Q: Why is the stopping potential negative in reality, but we use a positive value in the equation?

The stopping potential is indeed the negative potential applied to the anode with respect to the cathode to repel the electrons. However, in the energy conservation equation $K_{\text{max}} = eV_0$, the term $V_0$ represents the absolute magnitude of this potential difference, which is a positive quantity.

Question

The stopping potential for electrons emitted from a photosensitive surface illuminated by light of wavelength $491	ext{ nm}$ is $0.710	ext{ V}$. When the incident wavelength is changed to a new value, the stopping potential is $1.43	ext{ V}$. The new wavelength is

Options:

(A) $309	ext{ nm}$

(B) $329	ext{ nm}$

(C) $382	ext{ nm}$

(D) $400	ext{ nm}$

Sigma Percentile · Accepted Answer

### The Magic of the Photoelectric Effect

Imagine you are standing in front of a microscopic catapult system. Instead of rocks, this catapult launches electrons, and instead of a mechanical lever, it is triggered by pure light. This is the essence of the photoelectric effect, a phenomenon that completely revolutionized our understanding of physics and earned Albert Einstein his Nobel Prize.

When light of a specific wavelength strikes a photosensitive metal surface, it acts like a stream of tiny energy packets called photons. If a photon has enough energy, it can knock an electron completely out of the metal. However, the metal doesn't let go easily; it holds onto its electrons with a certain binding energy known as the **work function** ($\phi_0$).

### The Stopping Potential: A Uphill Battle

Once the electron breaks free, any leftover energy from the photon becomes the electron's kinetic energy. Some electrons barely make it out, while others fly out with maximum kinetic energy ($K_{	ext{max}}$).

To measure this maximum kinetic energy, physicists use a clever trick: they set up an electric field that pushes *against* the escaping electrons. They keep cranking up the voltage until even the fastest, most energetic electron is stopped dead in its tracks and turned around. This specific voltage is called the **stopping potential** ($V$).

Mathematically, the energy required to climb this electrical "hill" is $eV$, which must perfectly match the electron's maximum kinetic energy. This brings us to Einstein's elegant master equation:

$$E_{	ext{photon}} = \phi_0 + K_{	ext{max}}$$

$$\frac{hc}{\lambda} = \phi_0 + eV$$

### Setting Up the Mathematical Duel

In our specific problem, we are given two distinct scenarios for the same metal surface. Because it's the same metal, the work function $\phi_0$ remains a constant, hidden variable.

To make our calculations incredibly smooth, we use a beloved shortcut in modern physics: $hc \approx 1240 	ext{ eV}\cdot	ext{nm}$. This allows us to plug in the wavelength directly in nanometers and get the energy in electron-volts.

Let's write down the equations for our two cases.

**Case 1:** The incident wavelength $\lambda_1$ is $491 	ext{ nm}$, and the stopping potential $V_1$ is $0.710 	ext{ V}$.

$$\frac{1240}{491} = \phi_0 + 0.710$$

**Case 2:** The incident wavelength $\lambda_2$ is unknown, but the new stopping potential $V_2$ is $1.43 	ext{ V}$.

$$\frac{1240}{\lambda_2} = \phi_0 + 1.43$$

### The Art of Elimination

We have a system of two equations, but we don't care about the work function $\phi_0$. The most elegant way forward is to eliminate it entirely. By subtracting the first equation from the second, $\phi_0$ vanishes into thin air!

$$\left( \frac{1240}{\lambda_2} ight) - \left( \frac{1240}{491} ight) = (\phi_0 + 1.43) - (\phi_0 + 0.710)$$

Factoring out the $1240$ on the left side, we get:

$$1240 \left( \frac{1}{\lambda_2} - \frac{1}{491} ight) = 0.72$$

### The Final Calculation

Now, it's just a matter of careful arithmetic. Let's isolate the term containing our unknown wavelength, $\lambda_2$.

$$\frac{1}{\lambda_2} - \frac{1}{491} = \frac{0.72}{1240}$$

Calculating the fractions:

$$\frac{1}{\lambda_2} - 0.00204 \approx 0.00058$$

Moving the constant to the right side:

$$\frac{1}{\lambda_2} \approx 0.00058 + 0.00204 = 0.00262 	ext{ nm}^{-1}$$

Finally, we take the reciprocal to find the wavelength:

$$\lambda_2 = \frac{1}{0.00262} \approx 381.7 	ext{ nm}$$

Rounding to the nearest whole number, we get **$382 	ext{ nm}$**.

Notice the beautiful physical consistency here: to achieve a higher stopping potential ($1.43 	ext{ V}$ compared to $0.710 	ext{ V}$), the emitted electrons must have had more kinetic energy. This means the incident photons must have packed a bigger punch, which corresponds to a **shorter wavelength**. The math perfectly mirrors the physical reality!

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Visualized Solution

\text{Photoelectric Setup}

\text{Einstein's Photoelectric Equation}

\text{Setting up the Cases}

\text{Eliminating Work Function}

\text{Solving for } \lambda_2

\text{Final Wavelength}

\text{Conceptual Check}

The Sigma Insight: Photoelectric Effect

The Magic of the Photoelectric Effect