Two metallic plates A and B each of area 5 x 10^-4 m^2, are placed parallel to each other at separat

Q: Why does plate A become positively charged?

When light strikes plate A, it ejects photoelectrons. Since electrons carry a negative charge, losing them leaves plate A with a net positive charge.

Q: Why is the electric field directed from B to A?

Electric field lines originate from positive charges and terminate on negative charges. Since plate B has a larger positive charge than plate A, the net electric field points away from B towards A.

Q: Why do the electrons accelerate towards plate B?

Electrons are negatively charged, so they experience an electric force in the direction opposite to the electric field. Since the field points from B to A, the force pushes the electrons towards B, accelerating them.

Q: How do we calculate the electric field between two charged plates?

For two large parallel plates with charges $q_A$ and $q_B$, the electric field in the region between them is given by $E = \frac{q_B - q_A}{2A\varepsilon_0}$. This comes from the superposition of the electric fields produced by each plate individually.

Q: Why do we add the work done by the electric field to the initial kinetic energy?

According to the work-energy theorem, the total work done on a particle equals its change in kinetic energy. The electric field does positive work on the electrons, increasing their kinetic energy as they move from A to B.

Question

Two metallic plates $A$ and $B$ each of area $5 	imes 10^{-4} 	ext{ m}^2$, are placed parallel to each other at separation of $1 	ext{ cm}$. Plate $B$ carries a positive charge of $33.7 	imes 10^{-12} 	ext{ C}$. A monochromatic beam of light, with photons of energy $5 	ext{ eV}$ each, starts falling on plate $A$ at $t = 0$ so that $10^{16}$ photons fall on it per square metre per second. Assume that one photoelectron is emitted for every $10^6$ incident photons. Also assume that all the emitted photoelectrons are collected by plate $B$ and the work function of plate $A$ remains constant at the value $2 	ext{ eV}$.
Determine
(a) the number of photoelectrons emitted up to $t = 10 	ext{ s}$,
(b) the magnitude of the electric field between the plates $A$ and $B$ at $t = 10 	ext{ s}$ and
(c) the kinetic energy of the most energetic photoelectrons emitted at $t = 10 	ext{ s}$ when it reaches plate $B$.
Neglect the time taken by the photoelectron to reach plate $B$.
(Take $\varepsilon_0 = 8.85 	imes 10^{-12} 	ext{ C}^2 / 	ext{N-m}^2$).

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine two parallel metallic plates, $A$ and $B$, separated by a small distance of $1 	ext{ cm}$. Plate $B$ is initially given a positive charge of $33.7 	imes 10^{-12} 	ext{ C}$, while plate $A$ is neutral. Suddenly, a monochromatic beam of light with a photon energy of $5 	ext{ eV}$ starts illuminating plate $A$.

This light isn't just shining; it's knocking electrons out of plate $A$ due to the photoelectric effect. We are given the rate at which photons strike the plate: $10^{16}$ photons per square meter per second. However, not every photon is successful in ejecting an electron. The quantum efficiency is $1$ in $10^6$, meaning only one photoelectron is emitted for every million incident photons.

### The Photoelectron Count
First, let's determine exactly how many photoelectrons are emitted in the first $10 	ext{ s}$. We start by calculating the total number of photons striking plate $A$.

$$N_p = (	ext{rate}) 	imes 	ext{Area} 	imes 	ext{time}$$
$$N_p = 10^{16} 	imes (5 	imes 10^{-4}) 	imes 10 = 5 	imes 10^{13} 	ext{ photons}$$

Applying the quantum efficiency, the number of emitted photoelectrons $n$ is:

$$n = \frac{5 	imes 10^{13}}{10^6} = 5 	imes 10^7 	ext{ electrons}$$

### Charge Redistribution and Electric Field
When plate $A$ emits these $5 	imes 10^7$ electrons, it loses negative charge, which means it acquires a net positive charge.

$$q_A = +ne = (5 	imes 10^7) 	imes (1.6 	imes 10^{-19}) = 8.0 	imes 10^{-12} 	ext{ C}$$

These electrons travel across the gap and are collected by plate $B$. Plate $B$ initially had a positive charge, but absorbing these electrons reduces its net positive charge.

$$q_B = 33.7 	imes 10^{-12} - 8.0 	imes 10^{-12} = 25.7 	imes 10^{-12} 	ext{ C}$$

Now, we have two parallel plates with different positive charges. The electric field $E$ in the region between them is determined by the difference in their charges:

$$E = \frac{q_B - q_A}{2A\varepsilon_0}$$

Substituting the values:

$$E = \frac{(25.7 - 8.0) 	imes 10^{-12}}{2 	imes (5 	imes 10^{-4}) 	imes (8.85 	imes 10^{-12})} = \frac{17.7 	imes 10^{-12}}{88.5 	imes 10^{-16}} = 2 	imes 10^3 	ext{ N/C}$$

Since plate $B$ has a larger positive charge, the electric field points from $B$ to $A$.

### The Final Kinetic Energy
To find the kinetic energy of the electrons when they crash into plate $B$, we first need their initial kinetic energy as they leave plate $A$. Using Einstein's photoelectric equation:

$$K_{\max} = h
u - W = 5 	ext{ eV} - 2 	ext{ eV} = 3 	ext{ eV}$$

As these electrons travel towards plate $B$, they find themselves in an electric field pointing from $B$ to $A$. Because electrons are negatively charged, they experience a force in the opposite direction of the field—meaning they are pushed towards plate $B$! This force accelerates them, doing positive work.

The work done by the electric field is:

$$W_E = eEd = e(2 	imes 10^3 	ext{ V/m})(10^{-2} 	ext{ m}) = 20 	ext{ eV}$$

By the work-energy theorem, this work adds to their initial kinetic energy. The final kinetic energy $K_f$ of the most energetic electrons when they reach plate $B$ is:

$$K_f = K_{\max} + W_E = 3 	ext{ eV} + 20 	ext{ eV} = \mathbf{23 	ext{ eV}}$$

Visualized Solution

$N_{p} = rate \times A \times t$

$q_{A} = + n e, q_{B} = q_{B 0} - n e$

$E = \frac{q _{B} - q _{A}}{2 A ε _{0}}$

$K_{m a x} = h ν - W$

$K_{f} = K_{m a x} + W_{E}$

The Sigma Insight: Photoelectric Effect

Analyzing the Setup Imagine two parallel metallic plates, $A$ and $B$ , separated by a small distance of $1 cm$

The Photoelectron Count First, let's determine exactly how many photoelectrons are emitted in the first $10 s$

Charge Redistribution and Electric Field

The Final Kinetic Energy To find the kinetic energy of the electrons when they crash into plate $B$ , we first need their initial kinetic energy as they leave plate $A$

Similar Questions

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Np​=rate×A×t

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } qA​=+ne,qB​=qB0​−ne

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } E=2Aε0​qB​−qA​​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kmax​=hν−W

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Kf​=Kmax​+WE​