A metal plate of area 1 x 10^-4 m^2 is illuminated by a radiation of intensity 16 mW/m^2. The work f

Q: Why do we convert the photon energy from eV to Joules when calculating the number of photons?

Power is given in Watts, which is Joules per second. To maintain dimensional consistency and find the correct number of photons per second, the energy of a single photon must also be in Joules.

Q: What does the 10% efficiency physically mean in this context?

It means that out of every 100 photons that strike the metal surface, only 10 successfully transfer their energy to an electron in a way that allows it to escape. The other 90 photons might be reflected, scattered, or their energy might just heat up the metal lattice without ejecting an electron.

Q: Why is it called the *maximum* kinetic energy?

Electrons are bound at different depths within the metal. The work function $\phi$ is the *minimum* energy required to eject an electron from the very surface. Electrons deeper inside require more energy to reach the surface, so they emerge with less kinetic energy. The ones at the surface emerge with the *maximum* possible kinetic energy.

Q: If we increased the intensity of the light, would the maximum kinetic energy change?

No! Increasing intensity means more photons are hitting the plate per second, which would increase the *number* of emitted electrons. However, the energy of each individual photon remains $10 \text{ eV}$, so the maximum kinetic energy of any single electron remains $5 \text{ eV}$.

Q: What would happen if the incident photon energy was $4 \text{ eV}$ instead of $10 \text{ eV}$?

Since $4 \text{ eV}$ is less than the work function of $5 \text{ eV}$, no electrons would be emitted at all, regardless of how high the intensity is. This is the threshold phenomenon that classical wave theory couldn't explain!

Question

A metal plate of area $1 	imes 10^{-4} 	ext{ m}^2$ is illuminated by a radiation of intensity $16 	ext{ mW/m}^2$. The work function of the metal is $5 	ext{ eV}$. The energy of the incident photons is $10 	ext{ eV}$ and only $10\%$ of it produces photoelectrons. The number of emitted photoelectrons per second and their maximum energy, respectively will be (Take, $1 	ext{ eV} = 1.6 	imes 10^{-19} 	ext{ J}$)

Options:

(A) $10^{11}$ and $5 	ext{ eV}$

(B) $10^{12}$ and $5 	ext{ eV}$

(C) $10^{10}$ and $5 	ext{ eV}$

(D) $10^{14}$ and $10 	ext{ eV}$

Sigma Percentile · Accepted Answer

### The Photoelectric Setup: Setting the Stage

Imagine a metal plate resting in a vacuum, suddenly bombarded by a relentless rain of light. This isn't just any light; it's a stream of high-energy photons, each carrying exactly $10 	ext{ eV}$ of energy. The metal plate itself has a work function, $\phi = 5 	ext{ eV}$, which acts like a toll booth. An electron must pay this exact amount of energy just to break free from the metal's surface.

Our mission is twofold: we need to find out how energetic these escaping electrons are, and exactly how many of them manage to escape every single second.

### Einstein's Master Equation: The Energy Balance

Let's tackle the kinetic energy first. This is where Albert Einstein's Nobel Prize-winning photoelectric equation comes into play. The law of conservation of energy dictates that the energy of the incoming photon is split into two parts: paying the 'toll' (the work function) and giving the electron its kinetic energy.

$$K_{	ext{max}} = E - \phi$$

Substituting our known values is straightforward:

$$K_{	ext{max}} = 10 	ext{ eV} - 5 	ext{ eV} = 5 	ext{ eV}$$

So, the most energetic electrons will fly off the plate with a kinetic energy of $5 	ext{ eV}$. That's half of our problem solved!

### Calculating the Photon Flux: The Rain of Light

Now, to find the number of electrons, we must first understand the sheer volume of light hitting the plate. We are given the intensity of the light, $I = 16 	ext{ mW/m}^2$, and the area of the plate, $A = 1 	imes 10^{-4} 	ext{ m}^2$.

Intensity is simply power per unit area. Therefore, the total power (energy per second) striking the plate is:

$$P = I 	imes A$$

$$P = (16 	imes 10^{-3} 	ext{ W/m}^2) 	imes (10^{-4} 	ext{ m}^2) = 16 	imes 10^{-7} 	ext{ W}$$

This means $16 	imes 10^{-7} 	ext{ Joules}$ of energy hit the plate every second. But this energy arrives in discrete packets—photons. To find the total number of incident photons per second ($n_p$), we divide the total power by the energy of a single photon.

$$n_p = \frac{P}{E}$$

**Crucial Step:** We must ensure our units match! The power is in Joules per second, so the photon energy must be converted from electron-volts to Joules.

$$E = 10 	ext{ eV} = 10 	imes 1.6 	imes 10^{-19} 	ext{ J} = 16 	imes 10^{-19} 	ext{ J}$$

Now, we can find the photon flux:

$$n_p = \frac{16 	imes 10^{-7}}{16 	imes 10^{-19}} = 10^{12} 	ext{ photons/second}$$

### The Efficiency Catch: The Final Tally

A trillion photons are striking the plate every second! However, the physical world is rarely perfectly efficient. The problem states that only $10\%$ of these photons successfully eject an electron. The rest might be reflected or simply heat the metal.

Therefore, the number of emitted photoelectrons per second ($n_e$) is:

$$n_e = 10\% 	ext{ of } n_p$$

$$n_e = \frac{10}{100} 	imes 10^{12} = 10^{11} 	ext{ electrons/second}$$

And there we have it! The metal plate emits $10^{11}$ electrons every second, and the fastest among them zip away with a kinetic energy of $5 	ext{ eV}$.

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Final Answer

The Way Forward

The Sigma Insight: Photoelectric Effect

The Photoelectric Setup

Einstein's Master Equation

Calculating the Photon Flux

The Efficiency Catch

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