In a historical experiment to determine Planck's constant, a metal surface was irradiated with light

Q: Why didn't we use the third data point?

Two data points are sufficient to solve for the two unknowns ($h$ and $\phi$). The third point is consistent with the first two and can be used to verify the result.

Q: What is the significance of the stopping potential?

The stopping potential $V_0$ is the exact voltage needed to stop the most energetic photoelectrons, meaning $eV_0 = K_{\max}$.

Q: Why does the stopping potential decrease as wavelength increases?

A longer wavelength means lower energy photons. Lower energy photons give less kinetic energy to the emitted electrons, requiring a smaller stopping potential.

Q: Can we find the work function of the metal from this data?

Yes! Once you have $h$, you can substitute it back into either of the initial equations to solve for $\phi$.

Q: Why do we need to convert micrometers to meters?

To ensure all units are in the standard SI system, which is necessary when using the given values for $c$ (m/s) and $e$ (Coulombs) to get $h$ in Joule-seconds.

Question

In a historical experiment to determine Planck's constant, a metal surface was irradiated with light of different wavelengths. The emitted photoelectron energies were measured by applying a stopping potential. The relevant data for the wavelength ($\lambda$) of incident light and the corresponding stopping potential ($V_0$) are given below:

\begin{array}{cc}
\hline
\lambda 	ext{ (}\mu	ext{m)} & V_0 	ext{ (Volt)} \
\hline
0.3 & 2.0 \
0.4 & 1.0 \
0.5 & 0.4 \
\hline
\end{array}

Given that $c=3	imes 10^8 	ext{ ms}^{-1}$ and $e=1.6	imes 10^{-19}	ext{ C}$, Planck's constant (in units of J-s) found from such an experiment is

Options:

(A) $6.0	imes 10^{-34}$

(B) $6.4 	imes 10^{-34}$

(C) $6.6	imes 10^{-34}$

(D) $6.8	imes 10^{-34}$

Sigma Percentile · Accepted Answer

### The Photoelectric Effect and Planck's Constant

Imagine shining a light on a piece of metal and watching electrons pop out. This is the **photoelectric effect**, a phenomenon that baffled classical physicists but was elegantly explained by Albert Einstein. He proposed that light is made of tiny packets of energy called photons. The energy of each photon is given by $E = \frac{hc}{\lambda}$, where $h$ is Planck's constant, $c$ is the speed of light, and $\lambda$ is the wavelength.

### Einstein's Master Equation

When a photon hits the metal, it transfers its energy to an electron. Some of this energy is used to break the electron free from the metal's surface—this is called the **work function** ($\phi$). The leftover energy becomes the electron's kinetic energy. If we apply a reverse voltage, called the **stopping potential** ($V_0$), we can stop even the fastest electrons. This gives us Einstein's photoelectric equation:

$$\frac{hc}{\lambda} - \phi = eV_0$$

### Using Data to Eliminate the Unknown

In our experiment, we have two unknowns: Planck's constant ($h$) and the work function ($\phi$). However, we are given multiple data points. By plugging in the values for two different wavelengths, we get a system of two equations:

For $\lambda_1 = 0.3\,\mu	ext{m}$ and $V_{01} = 2.0	ext{ V}$:
$$\frac{hc}{0.3 	imes 10^{-6}} - \phi = e(2.0)$$

For $\lambda_2 = 0.4\,\mu	ext{m}$ and $V_{02} = 1.0	ext{ V}$:
$$\frac{hc}{0.4 	imes 10^{-6}} - \phi = e(1.0)$$

By subtracting the second equation from the first, the unknown work function $\phi$ is completely eliminated!

### The Final Calculation

Subtracting the equations gives:
$$hc 	imes 10^6 \left( \frac{1}{0.3} - \frac{1}{0.4} ight) = e(2.0 - 1.0)$$

Simplifying the terms inside the bracket:
$$hc 	imes 10^6 \left( \frac{10}{3} - \frac{10}{4} ight) = e$$
$$hc 	imes 10^6 \left( \frac{10}{12} ight) = e$$

Now, we just isolate $h$ and plug in the known values for $c$ ($3 	imes 10^8 	ext{ m/s}$) and $e$ ($1.6 	imes 10^{-19} 	ext{ C}$):
$$h = \frac{1.2 	imes 1.6 	imes 10^{-19}}{10^6 	imes 3 	imes 10^8}$$
$$h = 0.64 	imes 10^{-33} = 6.4 	imes 10^{-34} 	ext{ J-s}$$

This brilliant technique allows us to calculate one of the most fundamental constants of the universe using simple laboratory measurements!

Select Answer:

Visualized Solution

$Einstein’s Photoelectric Equation$

$Case 1: λ_{1} = 0.3 μ m$

$Case 2: λ_{2} = 0.4 μ m$

$Eliminating ϕ$

$Solving for h$

$Final Calculation$

The Sigma Insight: Photoelectric Effect

The Photoelectric Effect and Planck's Constant

Einstein's Master Equation

Using Data to Eliminate the Unknown

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; } Einstein’s Photoelectric Equation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 1: λ1​=0.3μm

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Case 2: λ2​=0.4μm

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

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

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