In a Frank-Hertz experiment, an electron of energy 5.6 eV passes through mercury vapour and emerges

Q: Why do we assume the entire lost energy goes into a single photon?

The question asks for the *minimum* wavelength. Since $\lambda = \frac{hc}{E}$, minimum wavelength corresponds to the maximum energy transition. This happens when the atom releases all the absorbed energy in a single jump.

Q: Why didn't the electron lose all its $5.6\text{ eV}$ energy?

This is the beauty of the Franck-Hertz experiment! Energy transfer in atomic collisions is quantized. The electron can only transfer energy in specific amounts that exactly match the energy level differences of the mercury atom.

Q: Why do we use $1240\text{ eV}\cdot\text{nm}$ for $hc$ instead of standard SI units?

Using $hc \approx 1240\text{ eV}\cdot\text{nm}$ (or $1242$) is a powerful shortcut in modern physics. It allows you to directly plug in energy in electron-volts (eV) and get the wavelength in nanometers (nm), saving you from tedious powers-of-10 calculations.

Q: Could the mercury atom emit multiple photons instead of one?

Yes, it could cascade down through intermediate energy levels, emitting multiple lower-energy photons. However, these would have longer wavelengths. We are looking for the minimum wavelength, which comes from the single largest energy jump.

Q: What does the Franck-Hertz experiment actually prove?

It provides direct experimental evidence for the quantization of atomic energy levels, brilliantly validating the core postulate of the Bohr model of the atom.

Question

In a Frank-Hertz experiment, an electron of energy $5.6	ext{ eV}$ passes through mercury vapour and emerges with an energy $0.7	ext{ eV}$. The minimum wavelength of photons emitted by mercury atoms is close to

Options:

(A) $250	ext{ nm}$

(B) $2020	ext{ nm}$

(C) $1700	ext{ nm}$

(D) $220	ext{ nm}$

Sigma Percentile · Accepted Answer

### The Setup: A Journey Through Mercury Vapour

Imagine you are observing a sealed chamber filled with mercury vapour.

We fire an electron into this chamber with an initial kinetic energy of $5.6 	ext{ eV}$.

After traversing the vapour, the electron emerges on the other side, but its energy has dropped to just $0.7 	ext{ eV}$.

### The Collision: Energy Transfer in Action

So, where did the missing energy go?

According to the law of conservation of energy, it couldn't have just vanished.

During its journey, the electron collided with a mercury atom and transferred a specific chunk of its kinetic energy to it.

This absorbed energy causes the mercury atom to jump from its ground state to a higher, excited energy level.

We can easily calculate this absorbed energy by finding the difference between the electron's initial and final energies.

$$ \Delta E = 5.6 	ext{ eV} - 0.7 	ext{ eV} = 4.9 	ext{ eV} $$

### The Emission: Calculating the Wavelength

Now, this excited mercury atom is unstable and won't stay in that high-energy state forever.

It will eventually de-excite, dropping back down to the ground state.

As it drops, it releases the stored $4.9 	ext{ eV}$ of energy in the form of a photon.

The question asks for the **minimum wavelength** of the emitted photons.

Since wavelength is inversely proportional to energy, the minimum wavelength corresponds to the maximum energy transition.

This happens when the atom releases the entire $4.9 	ext{ eV}$ in a single, massive jump.

To find the wavelength, we use the classic Planck-Einstein relation:

$$ \lambda = \frac{hc}{\Delta E} $$

To make our calculation lightning fast, we use the standard approximation $hc \approx 1240 	ext{ eV}\cdot	ext{nm}$.

$$ \lambda = \frac{1240 	ext{ eV}\cdot	ext{nm}}{4.9 	ext{ eV}} $$

$$ \lambda \approx 253 	ext{ nm} $$

Looking at our options, the closest value is **$250 	ext{ nm}$**.

### The Legacy: Proving Bohr Right

This isn't just any random calculation; this is the historic Franck-Hertz experiment!

Before this experiment, the idea that atoms had discrete, quantized energy levels was just a theoretical postulate in Bohr's model.

By showing that electrons only lose specific, quantized amounts of energy when colliding with atoms, Franck and Hertz provided the first direct experimental proof of quantum energy levels.

It is a beautiful demonstration of quantum mechanics in action!

Animated Solution for Physics - Atoms and Nuclei: In a Frank-Hertz experiment, an electron of energy $5.6 eV$ passes through mercury vapour and emerges with an energy $0.7 eV$ . The minimum wavelength of photons emitted by mercury atoms is close to

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Visualized Solution

$The Setup$

$Energy Transfer$

$Energy Absorbed$

$Photon Emission$

$Wavelength Formula$

$Calculation$

$Final Answer$

$Significance$

The Sigma Insight: Bohr's Atomic Model and Energy Levels

The Setup

The Collision

The Emission

The Legacy

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Visualized Solution

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

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

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

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

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

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

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

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