When light of a given wavelength is incident on a metallic surface, the minimum potential needed to

Q: Why did we completely ignore the 'intensity half' part of the question?

Intensity refers to the number of photons striking the surface per second. While halving the intensity halves the number of emitted electrons (saturation current), it does not change the energy of individual photons. Therefore, the maximum kinetic energy and the stopping potential remain unaffected.

Q: How did $\frac{hc}{e}$ magically become $\text{eV m}$?

By definition, $1 \text{ eV}$ is the energy gained by an electron when accelerated through $1 \text{ Volt}$, which equals $e \text{ Joules}$ (where $e = 1.6 \times 10^{-19}$). Therefore, dividing Joules by $e$ converts the unit directly into electron-volts ($\text{eV}$).

Q: Why did we subtract the two equations instead of dividing them?

Subtracting the equations is the most efficient way to eliminate the work function ($\phi$), because $\phi$ is a constant property of the metal and appears as a linear additive term in both equations.

Q: What if the second stopping potential was given as a negative value?

Stopping potential is inherently a negative potential applied to the collector plate. However, in the photoelectric equation $K_{\max} = eV_s$, we use the magnitude of the stopping potential to represent the kinetic energy.

Q: Can the work function $\phi$ ever be negative?

No, the work function is the minimum energy required to pull an electron out of the metal's potential well. It is always a positive value.

Question

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0	ext{ V}$. This potential drops to $0.6	ext{ V}$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively?
[Take $\frac{hc}{e} = 1.24 	imes 10^{-6}	ext{ Jm C}^{-1}$]

Options:

(A) $1.72 	imes 10^{-7}	ext{ m}, 1.20	ext{ eV}$

(B) $1.72 	imes 10^{-7}	ext{ m}, 5.60	ext{ eV}$

(C) $3.78 	imes 10^{-7}	ext{ m}, 5.60	ext{ eV}$

(D) $3.78 	imes 10^{-7}	ext{ m}, 1.20	ext{ eV}$

Sigma Percentile · Accepted Answer

### The Photoelectric Setup

Imagine you are standing in a quantum laboratory. You have a pristine metal surface in front of you.

When a photon of light strikes this metal, it acts like a tiny billiard ball, transferring its energy to an electron.

If this energy is greater than the metal's **work function** ($\phi$), the electron is violently ejected.

To measure the maximum kinetic energy of these ejected electrons, we apply a negative **stopping potential** ($V_s$) to a collector plate.

When the fastest electron is just barely stopped from reaching the plate, we know its kinetic energy equals the electrical potential energy, $eV_s$.

### The Master Equation

The governing principle of this entire phenomenon is **Einstein's Photoelectric Equation**.

It elegantly states that the energy of the incident photon ($\frac{hc}{\lambda}$) is split into two parts.

One part pays the "toll fee" to escape the metal ($\phi$), and the rest becomes the electron's kinetic energy ($K_{\max}$).

$$ \frac{hc}{\lambda} = \phi + K_{\max} $$

Since $K_{\max} = eV_s$, we can rewrite this as:

$$ \frac{hc}{\lambda} = \phi + eV_s $$

### Analyzing the Two Cases

Let's look at the first scenario provided in the problem.

The incident light has a wavelength $\lambda$, and the stopping potential required is $6.0 	ext{ V}$.

Substituting these values into our master equation, we get our first relationship:

$$ \frac{hc}{\lambda} = \phi + 6.0 $$

Now, let's transition to the second scenario. The wavelength is increased to $4\lambda$.

The problem also mentions a trap: the intensity is halved. Does this matter?

**Absolutely not!** Intensity only dictates the *number* of photons per second, which changes the number of emitted electrons, but it has zero effect on their maximum kinetic energy.

The new stopping potential is $0.6 	ext{ V}$. Substituting this gives us our second equation:

$$ \frac{hc}{4\lambda} = \phi + 0.6 $$

### The Mathematical Execution

We now have a system of two linear equations with two variables: $\lambda$ and $\phi$.

Our immediate goal is to eliminate the work function ($\phi$) to find the energy of the photon.

We can achieve this by subtracting the second equation from the first.

$$ \left( \frac{hc}{\lambda} ight) - \left( \frac{hc}{4\lambda} ight) = (\phi + 6.0) - (\phi + 0.6) $$

Notice how beautifully $\phi$ cancels out! Simplifying the left side gives us $\frac{3hc}{4\lambda}$.

On the right side, $6.0 - 0.6$ simplifies to $5.4$.

$$ \frac{3hc}{4\lambda} = 5.4 $$

Now, let's isolate the term $\frac{hc}{\lambda}$.

Multiplying $5.4$ by $4$ and dividing by $3$ gives us the energy of the first photon.

$$ \frac{hc}{\lambda} = \frac{5.4 	imes 4}{3} = 7.2 	ext{ eV} $$

### Finding the Work Function

Awesome! We found the energy of the incident photon.

Now, let's plug this value back into our first equation to find the work function, $\phi$.

$$ 7.2 = \phi + 6.0 $$

Solving for $\phi$ is straightforward.

$$ \phi = 7.2 - 6.0 = 1.2 	ext{ eV} $$

The work function of the metal is **$1.2 	ext{ eV}$**.

### The Final Calculation

Finally, we need to determine the exact wavelength $\lambda$.

We know that $\frac{hc}{\lambda} = 7.2 	ext{ eV}$.

The problem provides a very convenient unit conversion: $\frac{hc}{e} = 1.24 	imes 10^{-6} 	ext{ Jm C}^{-1}$.

Since $1 	ext{ eV} = e 	ext{ Joules}$, this value is exactly equivalent to $1.24 	imes 10^{-6} 	ext{ eV m}$.

Let's substitute this into our wavelength equation.

$$ \lambda = \frac{hc}{7.2 	ext{ eV}} = \frac{1.24 	imes 10^{-6} 	ext{ eV m}}{7.2 	ext{ eV}} $$

Dividing these numbers gives us our final answer.

$$ \lambda = 1.72 	imes 10^{-7} 	ext{ m} $$

So, the wavelength of the first source is **$1.72 	imes 10^{-7} 	ext{ m}$**, and the work function is **$1.20 	ext{ eV}$**.

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Visualized Solution

\text{Photoelectric Effect Setup}

\text{Einstein's Photoelectric Equation}

\text{Case 1 Setup}

\text{Case 2 Setup}

\text{Eliminating } \phi

\text{Solving for } \frac{hc}{\lambda}

\text{Finding the Work Function}

\text{Unit Conversion for } \lambda

\text{Final Calculation}

The Sigma Insight: Photoelectric Effect

The Photoelectric Setup

The Master Equation

Analyzing the Two Cases

The Mathematical Execution

Finding the Work Function

The Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Photoelectric Effect Setup} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Einstein's Photoelectric Equation} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Case 2 Setup} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Solving for } \frac{hc}{\lambda} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Finding the Work Function} .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \text{Unit Conversion for } \lambda .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; }

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