What is the work function of the metal, if the light of wavelength 4000 A generates photoelectron of

Q: Why do we have to convert the wavelength from Angstroms to meters?

Because the speed of light ($c$) is given in meters per second and Planck's constant ($h$) involves Joules (which are based on kilograms, meters, and seconds). To ensure all units cancel correctly and give energy in Joules, wavelength must be in meters.

Q: Why do we divide by $1.6 \times 10^{-19}$ at the very end?

The standard SI unit for energy is the Joule. However, in atomic physics, energies are very small, so we often use the electron-volt (eV). One eV is the energy gained by an electron accelerated through a 1-volt potential difference, which equals $1.6 \times 10^{-19}$ Joules. Dividing by this factor converts Joules to eV.

Q: What exactly does the 'work function' mean physically?

Think of the work function as the 'toll fee' an electron must pay to escape the metal's surface. It is the absolute minimum energy required to break the attractive bonds holding the electron inside the metal lattice.

Q: Why is the kinetic energy in the formula often written as $KE_{max}$?

The formula assumes the electron absorbs the photon and escapes without any internal collisions. In reality, some electrons are deeper inside the metal and lose energy colliding with other atoms on their way out. Therefore, the formula calculates the *maximum* possible kinetic energy for an electron right at the surface.

Q: What would happen if the calculated photon energy was less than the work function?

If the incident photon energy ($E$) is less than the work function ($\Phi$), the electron doesn't have enough energy to pay the 'toll fee'. It cannot escape the metal, and the photoelectric effect simply does not occur, regardless of how intense or bright the light is.

Question

What is the work function of the metal, if the light of wavelength $4000 	ext{ \AA}$ generates photoelectron of velocity $6 	imes 10^5 	ext{ ms}^{-1}$ from it?
(Mass of electron $= 9 	imes 10^{-31} 	ext{ kg}$
Velocity of light $= 3 	imes 10^8 	ext{ ms}^{-1}$
Planck's constant $= 6.626 	imes 10^{-34} 	ext{ Js}$
Charge of electron $= 1.6 	imes 10^{-19} 	ext{ JeV}^{-1}$)

Options:

(A) $4.0 	ext{ eV}$

(B) $2.1 	ext{ eV}$

(C) $0.9 	ext{ eV}$

(D) $3.1 	ext{ eV}$

Sigma Percentile · Accepted Answer

### The Setup: Light Meets Metal

Imagine you are observing a microscopic game of pinball. A beam of light, acting as a stream of energetic particles called photons, strikes a metal surface. If a photon hits an electron with enough energy, it knocks the electron right out of the metal! This fascinating phenomenon is known as the **photoelectric effect**.

In our specific problem, we are given a light beam with a wavelength $\lambda = 4000 	ext{ \AA}$. When this light hits the metal, it ejects an electron that zooms away with a velocity $v = 6 	imes 10^5 	ext{ m/s}$. Our mission is to find the **work function** ($\Phi$) of the metal, which is the minimum energy required just to pull an electron free from the surface.

### The Master Equation: Einstein's Genius

To solve this, we turn to Albert Einstein's elegant photoelectric equation. It states that the total energy of the incident photon ($E$) is conserved and split into two parts: the energy used to overcome the work function ($\Phi$), and the remaining energy which becomes the kinetic energy ($KE$) of the ejected electron.

$$E = \Phi + KE$$

Rearranging this to solve for the work function, we get:

$$\Phi = E - KE$$

This means we need to calculate the photon's energy and the electron's kinetic energy separately, and then simply subtract them.

### Crunching the Numbers: Photon Energy

First, let's calculate the energy of the incoming photon using the formula $E = \frac{hc}{\lambda}$. We must be careful to convert the wavelength from Angstroms to meters by multiplying by $10^{-10}$.

$$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{4000 	imes 10^{-10}}$$

By carefully multiplying the constants and managing the powers of ten, we find:

$$E = \frac{19.878 	imes 10^{-26}}{4 	imes 10^{-7}} = 4.9695 	imes 10^{-19} 	ext{ J}$$

### The Electron's Escape: Kinetic Energy

Next, we determine the kinetic energy of the ejected electron using the classic mechanics formula $KE = \frac{1}{2}mv^2$. We plug in the mass of the electron and its given velocity.

$$KE = \frac{1}{2} 	imes 9 	imes 10^{-31} 	imes (6 	imes 10^5)^2$$

Squaring the velocity gives $36 	imes 10^{10}$. Multiplying everything together yields:

$$KE = \frac{1}{2} 	imes 9 	imes 10^{-31} 	imes 36 	imes 10^{10} = 162 	imes 10^{-21} 	ext{ J}$$

To make subtraction easier later, we adjust the decimal to match the power of ten of the photon's energy:

$$KE = 1.62 	imes 10^{-19} 	ext{ J}$$

### The Final Reveal: Calculating the Work Function

Now, we substitute our calculated values back into our rearranged photoelectric equation.

$$\Phi = 4.9695 	imes 10^{-19} - 1.62 	imes 10^{-19}$$

$$\Phi = 3.3495 	imes 10^{-19} 	ext{ J}$$

We have the work function in Joules, but our multiple-choice options are in electron-volts (eV). To convert Joules to eV, we divide by the charge of an electron ($1.6 	imes 10^{-19} 	ext{ C}$).

$$\Phi 	ext{ (in eV)} = \frac{3.3495 	imes 10^{-19}}{1.6 	imes 10^{-19}}$$

The powers of ten cancel out beautifully, leaving us with:

$$\Phi \approx 2.093 	ext{ eV}$$

Rounding to one decimal place, we get **$2.1 	ext{ eV}$**, which perfectly matches option (b). The physics holds true, and the math confirms it!

Select Answer:

Visualized Solution

\text{The Photoelectric Setup}

\text{Einstein's Photoelectric Equation}

\text{Energy of Incident Photon}

\text{Kinetic Energy of Electron}

\text{Calculating Work Function}

\text{Conversion to Electron-Volts}

\text{The Way Forward}

The Sigma Insight: Wave Particle Duality

The Setup

The Master Equation

Crunching the Numbers

The Electron's Escape

The Final Reveal

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