Assume that the de-Broglie wave associated with an electron can form a standing wave between the ato

Q: Why is the distance between nodes $\lambda/2$ and not $\lambda$?

A full wavelength $\lambda$ consists of one complete crest and one complete trough (two loops). Therefore, a single loop, which spans from one node to the next adjacent node, is exactly half of a wavelength, $\lambda/2$.

Q: How do we know the second wave has exactly $p+1$ loops?

The problem states that a similar standing wave forms at $2.5\text{ \AA}$ "but not for any intermediate value". This implies $2.5\text{ \AA}$ is the very next allowed state. Since the number of loops must be an integer, the next state after $p$ loops is $p+1$ loops.

Q: Why do we use the rest mass of the electron here?

The calculated kinetic energy is around $150\text{ eV}$, which is extremely small compared to the rest mass energy of an electron ($511,000\text{ eV}$). Therefore, relativistic effects are negligible, and we can safely use the classical rest mass $m = 9.1 \times 10^{-31}\text{ kg}$.

Q: Can the number of loops $p$ be a fraction?

No. For a standing wave to be stable with nodes at both ends, the total distance must be an integer multiple of half-wavelengths. If $p$ were a fraction, the wave wouldn't end on a node, and it would destructively interfere with itself.

Q: What does the least value of $d$ physically represent?

The least value of $d$ corresponds to the fundamental mode or the first harmonic of the standing wave, where only a single loop forms between the two boundaries.

Question

Assume that the de-Broglie wave associated with an electron can form a standing wave between the atoms arranged in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave is formed if the distance $d$ between the atoms of the array is $2	ext{ \AA}$. A similar standing wave is again formed if $d$ is increased to $2.5	ext{ \AA}$ but not for any intermediate value of $d$. Find the energy of the electron in eV and the least value of $d$ for which the standing wave of the type described above can form.

Sigma Percentile · Accepted Answer

### Visualizing the Standing Wave

Imagine an electron trapped in a one-dimensional array of atoms, bouncing back and forth. Because of its wave nature, it forms a **standing wave**, much like a plucked guitar string. For a standing wave to be stable, the ends must be nodes (points of zero amplitude).

The distance between any two adjacent nodes in a standing wave is exactly half of its wavelength, $\frac{\lambda}{2}$. If the total distance between the ends is $d$, and the wave forms $p$ loops, we can write:
$$p \left(\frac{\lambda}{2}ight) = d$$

### Setting Up the Equations

The problem states that a standing wave forms when the distance $d = 2	ext{ \AA}$. Let's assume this corresponds to $p$ loops:
$$p \left(\frac{\lambda}{2}ight) = 2	ext{ \AA}$$

When the distance is increased to $2.5	ext{ \AA}$, the *next* possible standing wave forms. Since no intermediate waves can form, this new wave must have exactly one more loop, meaning it has $p + 1$ loops:
$$(p + 1) \left(\frac{\lambda}{2}ight) = 2.5	ext{ \AA}$$

### Finding the Wavelength

We now have a simple system of two equations. By subtracting the first equation from the second, the $p$ terms cancel out beautifully:
$$(p + 1) \frac{\lambda}{2} - p \frac{\lambda}{2} = 2.5 - 2.0$$
$$\frac{\lambda}{2} = 0.5	ext{ \AA}$$

This immediately gives us the de-Broglie wavelength of the electron:
$$\lambda = 1.0	ext{ \AA} = 10^{-10}	ext{ m}$$

### Calculating the Kinetic Energy

Now that we have the wavelength, we can find the electron's kinetic energy using the de-Broglie relation:
$$\lambda = \frac{h}{p} = \frac{h}{\sqrt{2mK}}$$

Squaring both sides and rearranging for kinetic energy $K$:
$$K = \frac{h^2}{2m\lambda^2}$$

Let's plug in the standard constants ($h = 6.63 	imes 10^{-34}	ext{ J s}$, $m = 9.1 	imes 10^{-31}	ext{ kg}$):
$$K = \frac{(6.63 	imes 10^{-34})^2}{2 	imes (9.1 	imes 10^{-31}) 	imes (10^{-10})^2}	ext{ J}$$
$$K \approx 2.415 	imes 10^{-17}	ext{ J}$$

To convert this energy into electron-volts (eV), we divide by the elementary charge ($1.6 	imes 10^{-19}	ext{ C}$):
$$K = \frac{2.415 	imes 10^{-17}}{1.6 	imes 10^{-19}}	ext{ eV} \approx \mathbf{150.8	ext{ eV}}$$

### The Least Value of $d$

Finally, the question asks for the least value of $d$ for which a standing wave can form. The simplest possible standing wave consists of just a single loop ($p = 1$).

Therefore, the minimum distance is simply half a wavelength:
$$d_{	ext{min}} = 1 	imes \frac{\lambda}{2} = 0.5	ext{ \AA}$$

And there we have it! By understanding the geometry of standing waves, we unlocked the quantum properties of the electron.

Visualized Solution

$Standing Wave Condition$

$Next Standing Wave$

$Calculating Wavelength$

$Kinetic Energy Formula$

$Substituting Values$

$Calculating Energy in eV$

$Least Value of d$

The Sigma Insight: Matter Waves and de Broglie Relation

Visualizing the Standing Wave

Setting Up the Equations

Finding the Wavelength

Calculating the Kinetic Energy

The Least Value of $d$

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Visualized Solution

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

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

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

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

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

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

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