In a photoelectric effect set-up a point of light of power 3.2 x 10^-3 W emits monoenergetic photons

Q: Why do we use the solid angle approximation here?

Because the radius of the sphere ($8.0 \times 10^{-3}$ m) is much smaller than the distance from the source ($0.8$ m), the light rays hitting the sphere are nearly parallel, and the effective area intercepting the light is simply the cross-sectional area $\pi r^2$.

Q: Why does the emission of photoelectrons stop after some time?

As negatively charged electrons are ejected and swept away, the isolated sphere is left with a net positive charge. This creates a positive electric potential that attracts electrons back. Once this potential reaches the stopping potential, even the most energetic electrons cannot escape.

Q: How do we know the stopping potential?

The stopping potential $V_0$ is directly related to the maximum kinetic energy of the photoelectrons by the equation $K_{\max} = eV_0$. Since we found $K_{\max} = 2.0$ eV, the stopping potential is exactly $2.0$ V.

Q: Why do we use $12375$ to find the wavelength?

The value $12375$ (or sometimes $12400$) is a handy conversion factor for the product $hc$ when energy is in electron-volts (eV) and wavelength is in Angstroms ($\text{\AA}$). It saves a lot of time compared to using standard SI units!

Q: What does 'instantly swept away' mean in the problem?

It means that once an electron leaves the surface, it is removed from the vicinity of the sphere by some external mechanism, so we don't have to worry about a cloud of electrons forming a space charge around the sphere. We only need to consider the positive charge building up on the sphere itself.

Question

In a photoelectric effect set-up a point of light of power $3.2 	imes 10^{-3}$ W emits monoenergetic photons of energy $5.0$ eV. The source is located at a distance of $0.8$ m from the centre of a stationary metallic sphere of work function $3.0$ eV and of radius $8.0 	imes 10^{-3}$ m. The efficiency of photoelectrons emission is one for every $10^6$ incident photons. Assume that the sphere is isolated and initially neutral and that photoelectrons are instantly swept away after emission.
(a) Calculate the number of photoelectrons emitted per second.
(b) Find the ratio of the wavelength of incident light to the de-Broglie wavelength of the fastest photoelectrons emitted.
(c) It is observed that the photoelectrons emission stops at a certain time $t$ after the light source is switched on why ?
(d) Evaluate the time $t$.

Sigma Percentile · Accepted Answer

The photoelectric effect is one of the most beautiful phenomena in modern physics, bridging the gap between the wave and particle nature of light. In this problem, we are not just looking at a simple metal plate; we are dealing with an isolated metallic sphere. This adds a fascinating electrostatic twist to the standard photoelectric setup!

### Analyzing the Setup

Imagine a point source radiating light uniformly in all directions, like a tiny, powerful star. A small metallic sphere sits at a distance, intercepting a fraction of this light.

First, we need to understand how much light is actually hitting the sphere. The source has a power of $P = 3.2 	imes 10^{-3} 	ext{ W}$, and each photon carries an energy of $E_1 = 5.0 	ext{ eV}$.

By converting the photon energy into Joules, we can find the total number of photons emitted by the source every second:
$$n_1 = \frac{P}{E_1} = \frac{3.2 	imes 10^{-3}}{8.0 	imes 10^{-19}} = 4.0 	imes 10^{15} 	ext{ photons/s}$$

### The Solid Angle Interception

Now, the sphere is far away ($d = 0.8 	ext{ m}$) and quite small ($r = 8.0 	imes 10^{-3} 	ext{ m}$). It doesn't catch all the light! It only intercepts a fraction of the spherical wavefront.

We use the concept of solid angle to find the fraction of photons intercepted:
$$	ext{Fraction} = \frac{\pi r^2}{4\pi d^2}$$

Multiplying this fraction by the total photons emitted gives us the number of photons striking the sphere per second:
$$n_3 = n_1 	imes \frac{\pi r^2}{4\pi d^2} = 10^{11} 	ext{ photons/s}$$

### The Emission Efficiency

The problem states a harsh reality of the photoelectric effect: it's not very efficient. Only one in a million ($10^{-6}$) photons successfully ejects an electron.

So, the number of photoelectrons emitted per second is:
$$n = 10^{11} 	imes 10^{-6} = 10^5 	ext{ electrons/s}$$

This gives us the answer to part (a)!

### The Quantum Kinematics

For part (b), we need to dive into the quantum kinematics of the emitted electrons. Using Einstein's photoelectric equation, we find the maximum kinetic energy:
$$K_{\max} = E_1 - \Phi = 5.0 	ext{ eV} - 3.0 	ext{ eV} = 2.0 	ext{ eV}$$

Converting this to Joules ($3.2 	imes 10^{-19} 	ext{ J}$), we can calculate the de-Broglie wavelength of these fastest electrons:
$$\lambda_1 = \frac{h}{\sqrt{2mK_{\max}}} = 8.68 	imes 10^{-10} 	ext{ m} = 8.68 	ext{ \AA}$$

The wavelength of the incident light is simply:
$$\lambda_2 = \frac{12375}{5.0} = 2475 	ext{ \AA}$$

Taking the ratio $\frac{\lambda_2}{\lambda_1}$, we get **$285.1$**.

### The Electrostatic Trap

Part (c) asks a profound conceptual question. Why does the emission stop?

Visualize the isolated sphere. As it spits out negatively charged electrons, it loses negative charge. What happens to an isolated neutral object when it loses negative charge? It becomes **positively charged**!

This positive charge creates an electric potential that acts like a trap, pulling the escaping electrons back. As more electrons leave, this trap gets stronger. Eventually, the positive potential reaches the **stopping potential** ($V_0 = 2.0 	ext{ V}$), and even the fastest electrons are pulled back. The emission effectively stops.

### The Final Countdown

To find exactly when this happens (part d), we need to know how much charge is required to reach this $2.0 	ext{ V}$ potential.

Using the formula for the potential of a sphere:
$$V = \frac{1}{4\pi\varepsilon_0} \frac{q}{r}$$
$$2.0 = (9 	imes 10^9) \frac{q}{8.0 	imes 10^{-3}}$$

Solving for $q$, we find the required charge is $1.78 	imes 10^{-12} 	ext{ C}$.

Since we know the emission current ($i = n 	imes e = 1.6 	imes 10^{-14} 	ext{ A}$), the time $t$ is simply the total charge divided by the current:
$$t = \frac{q}{i} = \frac{1.78 	imes 10^{-12}}{1.6 	imes 10^{-14}} \approx 111 	ext{ s}$$

And there we have it! A beautiful interplay of quantum mechanics and classical electrostatics.

Visualized Solution

\text{Total Photons Emitted by Source}

\text{Photons Emitted per Second}

\text{Photons Incident on the Sphere}

\text{Photoelectrons Emitted per Second}

\text{Maximum Kinetic Energy}

\text{de-Broglie Wavelength of Electrons}

\text{Wavelength of Incident Light \& Ratio}

\text{Why does emission stop?}

\text{Charge Required to Stop Emission}

\text{Time to Stop Emission}

The Sigma Insight: Photoelectric Effect

Analyzing the Setup

The Solid Angle Interception

The Emission Efficiency

The Quantum Kinematics

The Electrostatic Trap

The Final Countdown

Similar Questions

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 $(\frac{h c}{e} = 1240 nm V)$