Photoelectric effect experiments are performed using three different metal plates p, q and r having

Q: Why does metal $p$ have the highest saturation current?

Saturation current is proportional to the number of photoelectrons emitted per second. Since metal $p$ has the lowest work function, all three wavelengths in the light beam have enough energy to eject electrons from it. Metals $q$ and $r$ only respond to two and one wavelengths respectively. Therefore, metal $p$ emits the most electrons, leading to the highest saturation current.

Q: Why do we only use the highest photon energy to calculate the stopping potential?

The stopping potential is the negative voltage required to stop the fastest-moving photoelectrons from reaching the anode. The fastest electrons are those ejected by the most energetic photons (shortest wavelength). Therefore, the maximum kinetic energy, and thus the stopping potential, is determined solely by the highest energy photon in the beam.

Q: What happens to the photons that don't have enough energy to overcome the work function?

If a photon's energy is less than the work function of the metal, it cannot free an electron. The energy is simply absorbed by the metal as thermal energy (heat), but no photoelectric emission occurs.

Q: Why are the stopping potentials negative?

The stopping potential is the retarding potential applied to the anode with respect to the cathode. It creates an electric field that opposes the motion of the emitted electrons. Since it's a retarding voltage, it's conventionally represented as a negative value on the $I-V$ graph.

Q: Does the intensity of the light affect the stopping potential?

No, the intensity of light only affects the number of photons incident per second, which in turn affects the saturation current. The stopping potential depends only on the frequency (or wavelength) of the incident light and the work function of the material.

Question

Photoelectric effect experiments are performed using three different metal plates $p, q$ and $r$ having work functions $\phi_p = 2.0$ eV, $\phi_q = 2.5$ eV and $\phi_r = 3.0$ eV, respectively. A light beam containing wavelengths of 550 nm, 450 nm and 350 nm with equal intensities illuminates each of the plates. The correct $I-V$ graph for the experiment is

Sigma Percentile · Accepted Answer

### Unraveling the Photoelectric Effect: A Tale of Three Metals

Imagine you are conducting a classic physics experiment. You have three different metal plates, $p$, $q$, and $r$, each with its own unique "energy toll" required to release an electron. This toll is known as the **work function** ($\phi$). We are given $\phi_p = 2.0$ eV, $\phi_q = 2.5$ eV, and $\phi_r = 3.0$ eV.

You shine a light beam on these plates. But this isn't just any light; it's a mixture of three distinct wavelengths: 550 nm, 450 nm, and 350 nm, all shining with equal intensity. Our goal is to predict the resulting $I-V$ (current vs. voltage) graph for each metal.

### The Energy of the Incoming Photons

To understand how these metals will react, we first need to find the energy of the incoming photons. We use the handy relation:

$$E = \frac{hc}{\lambda} = \frac{1240 	ext{ eV nm}}{\lambda}$$

Let's calculate the energy for each wavelength:

For $\lambda_1 = 550$ nm: 
$$E_1 = \frac{1240}{550} = 2.25 	ext{ eV}$$

For $\lambda_2 = 450$ nm: 
$$E_2 = \frac{1240}{450} = 2.75 	ext{ eV}$$

For $\lambda_3 = 350$ nm: 
$$E_3 = \frac{1240}{350} = 3.54 	ext{ eV}$$

### Who Gets to Emit Electrons?

For photoelectric emission to occur, the photon energy must be strictly greater than the work function of the metal ($E > \phi$). Let's see which metals respond to which photons.

**Metal $p$ ($\phi_p = 2.0$ eV):** 
Since $E_1, E_2,$ and $E_3$ are all greater than 2.0 eV, metal $p$ will emit photoelectrons for **all three** wavelengths.

**Metal $q$ ($\phi_q = 2.5$ eV):** 
The first photon (2.25 eV) doesn't have enough energy. So, only the 450 nm and 350 nm wavelengths ($E_2$ and $E_3$) will cause emission.

**Metal $r$ ($\phi_r = 3.0$ eV):** 
With a high work function of 3.0 eV, only the most energetic photon of 3.54 eV ($E_3$) can knock out electrons.

### Comparing Saturation Currents

The saturation current ($I_s$) depends on the total number of electrons emitted per second. Since the light intensities are equal, the metal that responds to the most wavelengths will emit the most electrons.

Because metal $p$ responds to all three wavelengths, it will have the highest saturation current. Metal $q$ will be in the middle, and metal $r$ will have the lowest saturation current.

$$	herefore I_{sp} > I_{sq} > I_{sr}$$

### Determining the Stopping Potentials

What about the stopping potential ($V_0$)? It is dictated by the fastest electrons, which are ejected by the highest energy photon ($E_3 = 3.54$ eV). We use Einstein's photoelectric equation:

$$|V_0| = E_{\max} - \phi$$

Let's calculate the magnitude of the stopping potential for each metal:

For metal $p$: 
$$|V_{0p}| = 3.54 - 2.0 = 1.54 	ext{ V}$$

For metal $q$: 
$$|V_{0q}| = 3.54 - 2.5 = 1.04 	ext{ V}$$

For metal $r$: 
$$|V_{0r}| = 3.54 - 3.0 = 0.54 	ext{ V}$$

$$	herefore |V_{0p}| > |V_{0q}| > |V_{0r}|$$

### The Final Verdict

We are looking for a graph where metal $p$ has the most negative stopping potential and the highest saturation current. Conversely, metal $r$ must have the least negative stopping potential and the lowest saturation current.

When we examine the options, **Graph (a)** perfectly illustrates this behavior, showing three distinct curves that align flawlessly with our physical deductions!

Select Answer:

Visualized Solution

Visual Anchor

Photon Energy Formula

Energy of First Wavelength

Energy of Second Wavelength

Energy of Third Wavelength

Emission from Metal $p$

Emission from Metal $q$

Emission from Metal $r$

Comparing Saturation Currents

Stopping Potential Logic

Stopping Potential for $p$

Stopping Potential for $q$

Stopping Potential for $r$

Final Conclusion

The Sigma Insight: Photoelectric Effect

Unraveling the Photoelectric Effect

The Energy of the Incoming Photons

Who Gets to Emit Electrons?

Comparing Saturation Currents

Determining the Stopping Potentials

The Final Verdict

Similar Questions

In a photoelectric effect experiment, the graph of stopping potential $V$ versus reciprocal of wavelength $(1/ λ)$ obtained is shown in the figure. As the intensity of incident radiation is increased,

In a photoelectric experiment, the wavelength of the light incident on a metal is changed from $300 nm$ to $400 nm$ . The decrease in the stopping potential is close to $(\frac{h c}{e} = 1240 nm V)$

When the wavelength of radiation falling on a metal is changed from $500 nm$ to $200 nm$ , the maximum kinetic energy of the photoelectrons becomes three times larger. The work function of the metal is close to

The figure shows the variation of photocurrent with anode potential for a photosensitive surface for three different radiations. Let $I_{a}, I_{b}$ and $I_{c}$ be the intensities and $f_{a}, f_{b}$ and $f_{c}$ be the frequencies for the curves $a, b$ and $c$ respectively. Then :

Select Answer:

Visualized Solution

Visual Anchor

Photon Energy Formula

Energy of First Wavelength

Energy of Second Wavelength

Energy of Third Wavelength

Emission from Metal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p

Emission from Metal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Emission from Metal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Comparing Saturation Currents

Stopping Potential Logic

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Stopping Potential for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } q

Stopping Potential for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Final Conclusion

The Sigma Insight: Photoelectric Effect

Unraveling the Photoelectric Effect

The Energy of the Incoming Photons

Who Gets to Emit Electrons?

Comparing Saturation Currents

Determining the Stopping Potentials

The Final Verdict

Similar Questions

Emission from Metal $p$

Emission from Metal $q$

Emission from Metal $r$

Stopping Potential for $p$

Stopping Potential for $q$

Stopping Potential for $r$