Q.The photoelectric cut-off voltage in a certain experiment is $1.5\ \text{V}$. What is the maximum kinetic energy of photoelectrons emitted?

The key idea is that the cut-off (stopping) voltage $V_0$ directly measures the maximum kinetic energy of the emitted photoelectrons, because the stopping potential just barely brings the fastest electrons to rest.

Reasoning:

1. The stopping potential $V_0$ is the voltage that gives the most energetic photoelectrons exactly enough work to overcome their kinetic energy: $K_{\text{max}} = e V_0$.
2. Here $V_0 = 1.5\ \text{V}$ and $e = 1.6 \times 10^{-19}\ \text{C}$.
3. So $K_{\text{max}} = (1.6 \times 10^{-19})(1.5) = 2.4 \times 10^{-19}\ \text{J}$.

The photoelectric effect is one of those rare experiments where a single measurement — the cut-off (or stopping) voltage — directly gives you the maximum kinetic energy of the emitted electrons. No need to know the work function or the incident light frequency. That’s the beauty of it.

Why does this work?

When you apply a reverse voltage between the emitter and collector, you create an electric field that opposes the motion of photoelectrons. The most energetic electrons — those with maximum kinetic energy — are the hardest to stop. The cut-off voltage $V_0$ is exactly the voltage needed to bring these fastest electrons to rest just as they reach the collector. At that point, the electrical potential energy gained ($eV_0$) equals the kinetic energy lost.

So the relation is direct:

where $e = 1.6 \times 10^{-19}\ \text{C}$ is the elementary charge.

Now let’s apply it.

1. Identify the given data.

   The cut-off voltage is $V_0 = 1.5\ \text{V}$.

2. Write the formula.

3. Compute in electronvolts (eV). Since $e \times 1\ \text{V} = 1\ \text{eV}$, the answer in eV is simply the numerical value of $V_0$:

4. Convert to joules (SI unit). Multiply by $e$:

The method is direct application of the photoelectric equation relating stopping potential to maximum kinetic energy.

Steps:

1. Recall the key relation: the stopping potential $V_0$ is the voltage that just stops the most energetic photoelectrons. The work done by the electric field in stopping them equals their maximum kinetic energy:

where $e$ is the elementary charge ($1.6 \times 10^{-19}\ \text{C}$).

2. You are given $V_0 = 1.5\ \text{V}$. Substitute directly:

3. Multiply:

Final answer:

$K_{\text{max}} = 2.4 \times 10^{-19}\ \text{J}$ (or $1.5\ \text{eV}$).

The most common mistake here is treating the cut-off voltage as if it were a potential difference that accelerates the electron, rather than a stopping potential. Students often multiply by the electron charge but then add or subtract something, or they forget that the unit "electronvolt" already accounts for the charge.

Mistake 1: Confusing cut-off voltage with accelerating voltage.

A cut-off voltage of $1.5\ \text{V}$ means you need to apply a retarding potential of $1.5\ \text{V}$ to just stop the fastest photoelectrons. The work done by the stopping potential equals the loss in kinetic energy: $e V_0 = K_{\text{max}}$. Some students think the kinetic energy is $eV_0$ plus the work function — that is wrong. The cut-off voltage directly gives the maximum kinetic energy; the work function is already accounted for in the fact that the voltage is the stopping value.

How to avoid: Remember the stopping condition: the electric field does negative work $-eV_0$ on the electron, reducing its kinetic energy to zero. So $K_{\text{max}} - eV_0 = 0$, hence $K_{\text{max}} = eV_0$. No extra terms.

Mistake 2: Forgetting to convert units properly.

The answer is often expected in electronvolts (eV) or joules. If the question asks for "maximum kinetic energy" without specifying units, give it in both eV and joules. A common error is to write $1.5\ \text{eV}$ but then incorrectly convert to joules (e.g., using $1.6 \times 10^{-19}$ but multiplying by $1.5$ twice, or using $1.6 \times 10^{-19}$ as if it were $1\ \text{eV}$ in volts).

How to avoid:

• In eV: $K_{\text{max}} = 1.5\ \text{eV}$ directly (since $V_0 = 1.5\ \text{V}$ and $e=1$ in eV units).
• In joules: $K_{\text{max}} = (1.6 \times 10^{-19}\ \text{C}) \times (1.5\ \text{V}) = 2.4 \times 10^{-19}\ \text{J}$. Do the multiplication once, carefully.

Mistake 3: Writing the formula incorrectly.

Some students write $K_{\text{max}} = hf - \phi$ and then try to relate $V_0$ to $hf$ or $\phi$ separately, getting tangled. They forget that $eV_0 = hf - \phi$ is the definition of the stopping potential. So $K_{\text{max}} = eV_0$ is a direct consequence, not a separate formula.

How to avoid: When you see "cut-off voltage" or "stopping potential", immediately write $K_{\text{max}} = eV_0$. That's the only relation you need for this question.

Mistake 4: Thinking the answer is $1.5\ \text{J}$ or $1.5\ \text{V}$.

A voltage is not an energy. The numerical value $1.5$ is the same, but the unit must be eV or J. Writing "$1.5$" without units loses marks.

How to avoid: Always attach the correct unit: electronvolt for atomic-scale problems, or joules if the problem context demands SI.

Final answer:

The maximum kinetic energy of the photoelectrons is $1.5\ \text{eV}$ (or $2.4 \times 10^{-19}\ \text{J}$).