In a photoelectric experiment, a parallel beam of monochromatic light with power of 200 W is inciden

Q: Why is the kinetic energy of the emitted electrons considered zero initially?

The problem states that the frequency of light is *just above* the threshold frequency. This means the photon energy is almost exactly equal to the work function, leaving negligible energy for the electron's initial kinetic energy.

Q: Why do we multiply by $1.6 \times 10^{-19}$ when calculating the number of photons?

The power is given in Watts (Joules per second), but the work function is in electron-volts (eV). To divide power by energy, they must be in the same units. Multiplying by $1.6 \times 10^{-19}$ converts eV to Joules.

Q: How did we get the formula $p = \sqrt{2mK}$?

Kinetic energy $K$ is $\frac{1}{2}mv^2$. Momentum $p$ is $mv$. If you square momentum, you get $p^2 = m^2v^2$. Dividing by $2m$ gives $\frac{p^2}{2m} = \frac{1}{2}mv^2 = K$. Rearranging this gives $p = \sqrt{2mK}$.

Q: Why is the force simply $N_e \times p$?

Force is the rate of change of momentum ($\frac{dp}{dt}$). Each electron brings a momentum $p$ and is absorbed, so its final momentum is zero. The change in momentum per electron is $p$. If $N_e$ electrons hit per second, the total change in momentum per second is $N_e \times p$.

Q: What would happen if the anode reflected the electrons perfectly instead of absorbing them?

If the electrons were perfectly reflected, their momentum would change from $p$ to $-p$, making the change in momentum $2p$ per electron. The force on the anode would then be exactly double, i.e., $F = 2 N_e p$.

Question

In a photoelectric experiment, a parallel beam of monochromatic light with power of $200	ext{ W}$ is incident on a perfectly absorbing cathode of work function $6.25	ext{ eV}$. The frequency of light is just above the threshold frequency, so that the photoelectrons are emitted with negligible kinetic energy. Assume that the photoelectron emission efficiency is $100\%$. A potential difference of $500	ext{ V}$ is applied between the cathode and the anode. All the emitted electrons are incident normally on the anode and are absorbed. The anode experiences a force $F = n 	imes 10^{-4}	ext{ N}$ due to the impact of the electrons. The value of $n$ is ........... . (Take mass of the electron, $m_e = 9 	imes 10^{-31}	ext{ kg}$ and $1	ext{ eV} = 1.6 	imes 10^{-19}	ext{ J}$)

Sigma Percentile · Accepted Answer

Have you ever imagined that light could push an object? While photons themselves carry momentum, in this fascinating JEE Advanced problem, we explore a two-step process: light knocks out electrons, and these electrons are then accelerated to smash into an anode, exerting a measurable physical force! Let's break down this beautiful interplay of quantum mechanics and classical kinematics.

### The Photon Bombardment

Our journey begins at the cathode. A powerful $200	ext{ W}$ monochromatic light beam is striking it. But how many photons are in this beam?

The problem gives us a crucial clue: the frequency of the light is *just above* the threshold frequency. According to Einstein's photoelectric equation, the energy of the incident photon is used to overcome the work function ($\phi$), and the rest becomes kinetic energy. If the frequency is just at the threshold, the kinetic energy is essentially zero. This means the energy of each photon is exactly equal to the work function: $E_{ph} = 6.25	ext{ eV}$.

To find the number of photons hitting the cathode per second ($N_{ph}$), we divide the total power by the energy of a single photon. Remember to convert electron-volts to Joules!
$$N_{ph} = \frac{P}{E_{ph}} = \frac{200}{6.25 	imes 1.6 	imes 10^{-19}}$$
$$N_{ph} = \frac{200}{10 	imes 10^{-19}} = 2 	imes 10^{20} 	ext{ photons/second}$$

### The Electron Avalanche

Next, we look at the efficiency of the cathode. The problem states that the photoelectron emission efficiency is $100\%$. In the quantum world, this is a perfect scenario: every single photon that hits the cathode successfully liberates exactly one electron.

Therefore, the number of electrons emitted per second ($N_e$) is exactly equal to the number of incident photons:
$$N_e = 2 	imes 10^{20} 	ext{ electrons/second}$$

### The Electric Accelerator

These $2 	imes 10^{20}$ electrons are just sitting at the cathode with negligible kinetic energy. But they don't stay there for long! A potential difference of $V = 500	ext{ V}$ is applied between the cathode and the anode.

This electric field acts as a particle accelerator. By the time an electron reaches the anode, it has acquired a kinetic energy equal to the work done by the electric field:
$$K = eV = 500	ext{ eV}$$

To find the force exerted on the anode, we need the momentum ($p$) of these electrons upon impact. The relationship between kinetic energy and momentum is a classic mechanics staple: $K = \frac{p^2}{2m_e}$, which gives us $p = \sqrt{2m_e K}$.

Let's plug in the numbers, converting the kinetic energy back to Joules:
$$p = \sqrt{2 	imes (9 	imes 10^{-31}) 	imes (1.6 	imes 10^{-19}) 	imes 500}$$
$$p = \sqrt{14400 	imes 10^{-50}} = 120 	imes 10^{-25} = 1.2 	imes 10^{-23} 	ext{ kg m/s}$$

### The Final Impact

Now for the grand finale. According to Newton's second law, force is the rate of change of momentum ($F = \frac{dp}{dt}$).

Every second, $N_e$ electrons crash into the anode. The problem explicitly states that the anode is *perfectly absorbing*. This means the electrons hit it and stop, transferring all their momentum to the anode. The change in momentum for each electron is simply $p$.

The total force $F$ is the total momentum transferred per second:
$$F = N_e 	imes p$$
$$F = (2 	imes 10^{20}) 	imes (1.2 	imes 10^{-23})$$
$$F = 2.4 	imes 10^{-3}	ext{ N}$$

The question asks for the force in the format $n 	imes 10^{-4}	ext{ N}$. Let's adjust our scientific notation:
$$F = 24 	imes 10^{-4}	ext{ N}$$

Comparing this to the given expression, we find our final answer:
**$n = 24$**

This problem is a brilliant demonstration of how macroscopic forces can arise from the microscopic bombardment of quantum particles. Always remember to track the energy and momentum at each stage of the journey!

Enter Numerical Value:

Visualized Solution

$Given Data$

$Energy of Incident Photon$

$Number of Electrons Emitted$

$Calculating N_{e}$

$Momentum of an Electron$

$Substituting Values for p$

$Calculating p$

$Force on Anode$

$Final Calculation$

$Reflection vs Absorption$

The Sigma Insight: Photoelectric Effect

The Photon Bombardment

The Electron Avalanche

The Electric Accelerator

The Final Impact

Similar Questions

In a photoelectric effect experiment, the threshold wavelength of light is $380 nm$ . If the wavelength of incident light is $260 nm$ , the maximum kinetic energy of emitted electrons will be Given, $E (in eV) = \frac{1237}{λ (in nm)}$

Enter Numerical Value:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Given Data

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Energy of Incident Photon

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Number of Electrons Emitted

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating Ne​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Momentum of an Electron

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values for p

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Force on Anode

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Calculation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Reflection vs Absorption