Q.The work function of caesium metal is $2.14\ \text{eV}$. When light of frequency $6 \times 10^{14}\ \text{Hz}$ is incident on the metal surface, photoemission of electrons occurs. What is the

Concept: Photon Energy — The energy of an incident photon is $E = h\nu$. Part of it goes into overcoming the work function $\phi$, and the remainder appears as the maximum kinetic energy $K_{\text{max}}$ of the photoelectron.

Step 1: Photon energy

$E = h\nu = (6.63 \times 10^{-34})(6 \times 10^{14}) = 3.978 \times 10^{-19}\ \text{J}$

Convert to eV: $E = \dfrac{3.978 \times 10^{-19}}{1.6 \times 10^{-19}} = 2.486\ \text{eV}$

Step 2: Maximum kinetic energy

$K_{\text{max}} = E - \phi = 2.486 - 2.14 = 0.346\ \text{eV}$

In joules: $K_{\text{max}} = 0.346 \times 1.6 \times 10^{-19} = 5.536 \times 10^{-20}\ \text{J}$

Step 3: Stopping potential

$V_0 = \dfrac{K_{\text{max}}}{e} = 0.346\ \text{V}$

Step 4: Maximum speed

$K_{\text{max}} = \frac12 m v_{\text{max}}^2 \implies v_{\text{max}} = \sqrt{\dfrac{2K_{\text{max}}}{m}}$

$v_{\text{max}} = \sqrt{\dfrac{2 \times 5.536 \times 10^{-20}}{9.1 \times 10^{-31}}} = 3.49 \times 10^5\ \text{m/s}$

The Core Idea: Photon Energy and the Photoelectric Effect

When light hits a metal surface, it behaves as a stream of particles — photons. Each photon carries a quantum of energy given by $E = hf$, where $h$ is Planck’s constant and $f$ is the frequency. For an electron to be ejected, the photon must supply enough energy to overcome the work function $\phi$ — the minimum energy needed to free an electron from the metal surface.

Any extra energy beyond $\phi$ appears as the maximum kinetic energy of the emitted electron. This is Einstein’s photoelectric equation:

The stopping potential $V_0$ is the voltage that just stops the most energetic electrons — it’s directly related to $K_{\text{max}}$ by $eV_0 = K_{\text{max}}$. And once we know $K_{\text{max}}$ in joules, the maximum speed follows from $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$.

Let’s apply this step by step.

Step 1: Find the photon energy

The incident light has frequency $f = 6 \times 10^{14}\ \text{Hz}$. Planck’s constant is $h = 6.63 \times 10^{-34}\ \text{J·s}$.

Photon energy in joules:

We’ll need this in electronvolts too. Since $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$:

Step 2: Maximum kinetic energy (part a)

Work function $\phi = 2.14\ \text{eV}$. Using Einstein’s equation:

Rounding to three significant figures (matching the given data):

Step 3: Stopping potential (part b)

The stopping potential $V_0$ satisfies $eV_0 = K_{\text{max}}$. Since $K_{\text{max}}$ is in eV, the numerical value of $V_0$ in volts is the same:

Step 4: Maximum speed (part c)

First convert $K_{\text{max}}$ to joules:

Electron mass $m = 9.1 \times 10^{-31}\ \text{kg}$. From $K_{\text{max}} = \frac{1}{2} m v_{\text{max}}^2$:

Calculate inside the square root:

Taking square root:

Method: Einstein's Photoelectric Equation

This problem is solved using Einstein's photoelectric equation, which states that the incident photon energy is used partly to overcome the work function and the remainder appears as the maximum kinetic energy of the emitted electron.

Step 1: Write down the given data

Step 2: Calculate the incident photon energy

Photon energy $E = hf$

Convert to eV:

Step 3: Find maximum kinetic energy (part a)

Einstein's photoelectric equation:

In joules:

Step 4: Find stopping potential (part b)

The stopping potential $V_0$ is related to $K_{\text{max}}$ by:

Step 5: Find maximum speed (part c)

Use kinetic energy in joules:

Final Answers

(a) Maximum kinetic energy: $0.346\ \text{eV}$ (or $5.54 \times 10^{-20}\ \text{J}$)

(b) Stopping potential: $0.346\ \text{V}$

(c) Maximum speed: $3.49 \times 10^5\ \text{m/s}$

Common Mistakes Students Make on This Photon Energy Problem

Mistake 1: Forgetting to convert units before using formulas

The work function is given in eV, but the Planck constant $h$ is usually taken in J·s ($6.63 \times 10^{-34}\ \text{J·s}$). Students often plug $2.14\ \text{eV}$ directly into $K_{\text{max}} = hf - \phi$ without converting everything to joules first.

How to avoid: Always check unit consistency. Convert the work function from eV to joules using $1\ \text{eV} = 1.6 \times 10^{-19}\ \text{J}$:

Now compute $hf$ in joules:

Then $K_{\text{max}} = 3.978 \times 10^{-19} - 3.424 \times 10^{-19} = 5.54 \times 10^{-20}\ \text{J}$.

Mistake 2: Confusing stopping potential with maximum kinetic energy

Students sometimes write $V_0 = K_{\text{max}}$ directly, forgetting that stopping potential is related by $eV_0 = K_{\text{max}}$.

How to avoid: Remember the definition: stopping potential is the voltage that just stops the most energetic electrons. The work done by the electric field ($eV_0$) equals the maximum kinetic energy lost. So:

Using $K_{\text{max}} = 5.54 \times 10^{-20}\ \text{J}$:

Mistake 3: Using the wrong mass for the electron in the speed calculation

Students sometimes use the mass of a proton or forget to square the speed properly in $K = \frac12 mv^2$.

How to avoid: The electron mass is $m_e = 9.1 \times 10^{-31}\ \text{kg}$. Rearranging:

Compute step by step:

• $2K_{\text{max}} = 1.108 \times 10^{-19}$
• Divide by $m_e$: $\frac{1.108 \times 10^{-19}}{9.1 \times 10^{-31}} = 1.218 \times 10^{11}$
• Take square root: $v_{\text{max}} = 3.49 \times 10^5\ \text{m/s}$

Mistake 4: Forgetting that photoemission requires $hf \geq \phi$

Some students attempt the problem even when the photon energy is below the work function, getting a negative kinetic energy.

How to avoid: First check if emission is possible. Here $hf = 3.978 \times 10^{-19}\ \text{J}$ and $\phi = 3.424 \times 10^{-19}\ \text{J}$, so $hf > \phi$ — emission occurs. If $hf < \phi$, simply state "no photoemission" and stop.

Mistake 5: Rounding intermediate values too aggressively

Rounding $K_{\text{max}}$ to $5.5 \times 10^{-20}\ \text{J}$ early can throw off the speed calculation significantly because of the square root.

How to avoid: Keep at least 3 significant figures throughout, and round only the final answer. Use $h = 6.63 \times 10^{-34}$ consistently (not $6.6 \times 10^{-34}$).

Final answers:

• (a) $K_{\text{max}} = \mathbf{5.54 \times 10^{-20}\ \text{J}}$ (or $0.346\ \text{eV}$)
• (b) $V_0 = \mathbf{0.346\ \text{V}}$
• (c) $v_{\text{max}} = \mathbf{3.49 \times 10^5\ \text{m/s}}$