A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which h

Q: Why does the stopping potential increase when we decrease the wavelength?

The energy of a photon is inversely proportional to its wavelength ($E = hc/\lambda$). A shorter wavelength means a higher energy photon. When this higher energy photon strikes the metal, it gives more kinetic energy to the ejected electron. To stop a faster electron, you need a stronger (higher magnitude) stopping potential.

Q: Can I just use $1240$ eV nm for the value of $hc$?

Yes, in many competitive exams like JEE and NEET, using $hc \approx 1240$ eV nm (or $1242$ eV nm) is a standard, time-saving approximation. However, if the problem explicitly provides values for $h$, $c$, and $e$, it is safer to calculate it to avoid slight rounding errors in the options.

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

Subtracting the equations is a mathematical shortcut. Since the metal is the same, the work function $\phi_0$ is identical in both cases. Subtracting eliminates $\phi_0$ immediately, saving you the time and effort of calculating an intermediate variable that isn't asked for.

Q: What would happen if we used a different metal for the second light source?

If you changed the metal, the work function $\phi_0$ would change. You would no longer be able to simply subtract the equations to eliminate it. You would need to know the specific work functions of both metals to solve the problem.

Q: Why is the stopping potential considered a 'negative' voltage?

Electrons are negatively charged. To stop them from reaching the collector plate, the collector plate must repel them. Therefore, the collector plate must be kept at a negative potential relative to the emitter plate. The 'stopping potential' value is usually stated as a positive magnitude, but physically it represents a retarding (negative) voltage.

Question

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm?

Options:

(A) 0.96 V

(B) 1.25 V

(C) 0.24 V

(D) 1.5 V

Sigma Percentile · Accepted Answer

### The Magic of the Photoelectric Effect

Imagine a world where light isn't just a continuous wave, but a stream of tiny, energetic packets called photons. When these photons strike a photosensitive metal surface, they can knock electrons right out of the metal! This phenomenon, known as the photoelectric effect, was brilliantly explained by Albert Einstein, earning him the Nobel Prize.

But these ejected electrons don't just sit there; they fly off with kinetic energy. To measure this energy, physicists use a clever trick: they apply a reverse voltage, called the **stopping potential** ($V_s$), to a collector plate. This negative voltage repels the electrons. When the voltage is just strong enough to stop even the fastest electrons from reaching the collector, we know exactly how much kinetic energy they had.

### Einstein's Master Equation

Einstein gave us a beautifully simple equation to describe this energy balance:

$$eV_s = \frac{hc}{\lambda} - \phi_0$$

Here, $e$ is the charge of an electron, $V_s$ is the stopping potential, $\frac{hc}{\lambda}$ is the energy of the incoming photon, and $\phi_0$ is the **work function**—the minimum energy required just to break the electron free from the metal's grip.

In our problem, we are dealing with the exact same metal surface, which means the work function $\phi_0$ is a constant. We are given two different scenarios with two different wavelengths of light.

**Case I:**
$$eV_1 = \frac{hc}{\lambda_1} - \phi_0$$

**Case II:**
$$eV_2 = \frac{hc}{\lambda_2} - \phi_0$$

### The Art of Elimination

We don't know the work function $\phi_0$, and frankly, we don't need to! By subtracting the first equation from the second, the pesky $\phi_0$ completely vanishes, leaving us with a clean relationship between the two stopping potentials:

$$eV_2 - eV_1 = \frac{hc}{\lambda_2} - \frac{hc}{\lambda_1}$$

Dividing everything by $e$, we isolate our target variable, $V_2$:

$$V_2 = V_1 + \frac{hc}{e} \left( \frac{1}{\lambda_2} - \frac{1}{\lambda_1} ight)$$

### Crunching the Numbers

Now comes the execution phase. We are given $V_1 = 0.48$ V, $\lambda_1 = 670.5$ nm, and $\lambda_2 = 474.6$ nm.

First, let's evaluate the constant term $\frac{hc}{e}$. Using standard values ($h = 6.63 	imes 10^{-34}$ J s, $c = 3 	imes 10^8$ m/s, and $e = 1.6 	imes 10^{-19}$ C), we get:

$$\frac{hc}{e} \approx 1243 	ext{ V nm}$$

*(Note: Many students use the approximation $1240$ eV nm or $1242$ eV nm for speed, but using $1243$ aligns perfectly with the exact constants provided in typical JEE problems).*

Substituting this into our equation:

$$V_2 = 0.48 + 1243 \left( \frac{1}{474.6} - \frac{1}{670.5} ight)$$

Taking the common denominator inside the bracket:

$$V_2 = 0.48 + 1243 \left( \frac{670.5 - 474.6}{474.6 	imes 670.5} ight)$$

$$V_2 = 0.48 + 1243 \left( \frac{195.9}{318219.3} ight)$$

$$V_2 = 0.48 + 1243 	imes 0.0006156$$

$$V_2 = 0.48 + 0.765$$

$$V_2 \approx 1.25 	ext{ V}$$

### The Physical Intuition

Does this answer make sense? Absolutely! We decreased the wavelength of the incident light from $670.5$ nm to $474.6$ nm. Because photon energy is inversely proportional to wavelength ($E = \frac{hc}{\lambda}$), the new photons pack a much bigger punch.

With more incoming energy, the ejected electrons fly out with greater kinetic energy. Naturally, it requires a stronger (more negative) stopping potential to halt them. Our calculated stopping potential increased from $0.48$ V to $1.25$ V, perfectly aligning with the physics of the universe!

Select Answer:

Visualized Solution

Photoelectric Effect Setup

Einstein's Photoelectric Equation

Eliminating Work Function

Substituting Values

Calculation

Final Calculation

Conclusion

The Sigma Insight: Photoelectric Effect

The Magic of the Photoelectric Effect

Einstein's Master Equation

The Art of Elimination

Crunching the Numbers

The Physical Intuition

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