Q.What is the de Broglie wavelength associated with

Concept: De Broglie Wavelength — every moving particle has a wavelength $\lambda = h/p$, where $h$ is Planck’s constant and $p$ is the linear momentum.

Step 1 — Formula

Step 2 — For the electron

$m_e = 9.11 \times 10^{-31}\ \text{kg}$, $v = 5.4 \times 10^{6}\ \text{m/s}$

Step 3 — For the ball

$m = 0.150\ \text{kg}$, $v = 30.0\ \text{m/s}$ …

The core idea here is wave-particle duality. Louis de Broglie proposed that every moving particle has an associated wavelength, just like a photon does. The wavelength is inversely proportional to the particle’s momentum — the heavier or faster the object, the shorter its wavelength. For macroscopic objects like a cricket ball, this wavelength is so tiny that it’s impossible to detect; for electrons, it’s comparable to atomic spacings, which is why electron microscopes work.

Let’s apply this to both cases.

(a) Electron moving at $5.4 \times 10^{6}\ \text{m/s}$

1. Identify the mass. The electron’s rest mass is $m_e = 9.11 \times 10^{-31}\ \text{kg}$. This is a standard value you must remember for such problems.

2. Compute momentum.

Multiply: $9.11 \times 5.4 = 49.194$, and $10^{-31} \times 10^{6} = 10^{-25}$.

So $p = 4.9194 \times 10^{-24}\ \text{kg·m/s}$.

3. Apply de Broglie relation.

Divide: $6.63 / 4.9194 \approx 1.347$, and $10^{-34} / 10^{-24} = 10^{-10}$.

So $\lambda \approx 1.35 \times 10^{-10}\ \text{m}$.

(b) Ball of mass $150\ \text{g}$ at $30.0\ \text{m/s}$

1. Convert mass to kg. $150\ \text{g} = 0.150\ \text{kg}$. This is a common slip — always use SI units.

2. Compute momentum.

3. Apply de Broglie relation.

$6.63 / 4.50 = 1.4733$, so

$\lambda \approx 1.47 \times 10^{-34}\ \text{m}$ …

Method: Direct Application of de Broglie's Relation

The de Broglie wavelength $\lambda$ for any moving particle is given by:

where $h = 6.63 \times 10^{-34}\ \text{J·s}$ is Planck's constant, $m$ is the mass in kg, and $v$ is the speed in m/s.

Steps

Step 1 — Identify the mass and speed for each object.

For (a), the electron: $m_e = 9.11 \times 10^{-31}\ \text{kg}$, $v = 5.4 \times 10^{6}\ \text{m/s}$.

For (b), the ball: $m = 150\ \text{g} = 0.150\ \text{kg}$, $v = 30.0\ \text{m/s}$.

Step 2 — Write the de Broglie relation and substitute.

For (a):

First compute the denominator:

$9.11 \times 10^{-31} \times 5.4 \times 10^{6} = 4.9194 \times 10^{-24}\ \text{kg·m/s}$.

Then:

Step 3 — Repeat for the ball.

For (b):

--- …

Common Mistakes with De Broglie Wavelength Problems

Students typically make the same few errors on this exact type of question. Here's what to watch out for.

1. Forgetting to convert mass to kilograms

The most frequent mistake. Mass must be in kg for the de Broglie formula $\lambda = \frac{h}{mv}$ to work with $h = 6.63 \times 10^{-34}\ \text{J·s}$.

In part (b), the ball's mass is given as $150\ \text{g}$. Many students plug in $150$ directly and get a wildly wrong answer.

How to avoid: Always check the units of every quantity before substituting. Convert grams to kilograms by dividing by 1000. Here, $150\ \text{g} = 0.150\ \text{kg}$.

2. Using the wrong value of Planck's constant

Some students use $h = 6.63 \times 10^{-34}$ but forget the exponent, or use $h = 6.63 \times 10^{34}$ (sign error). Others mistakenly use the reduced Planck constant $\hbar = \frac{h}{2\pi}$.

How to avoid: Memorise $h = 6.63 \times 10^{-34}\ \text{J·s}$ as a fixed fact. Write it down before starting the calculation. The de Broglie formula uses $h$, not $\hbar$.

3. Mixing up momentum with mass alone

The de Broglie wavelength is $\lambda = \frac{h}{p}$, where $p = mv$ is momentum. Some students write $\lambda = \frac{h}{m}$ or $\lambda = \frac{h}{v}$, leaving out velocity or mass.

How to avoid: Write the formula in full every time: $\lambda = \frac{h}{mv}$. Then substitute both mass and velocity together.

4. Unit errors in the final answer

After calculation, students often report the wavelength in metres when it should be in a more convenient unit like ångströms or nanometres. For an electron, the answer is typically on the order of $10^{-10}\ \text{m}$ (1 Å). For a macroscopic ball, it's unimaginably small — around $10^{-34}\ \text{m}$.

How to avoid: Before calculating, estimate the order of magnitude. An electron moving at $10^6\ \text{m/s}$ gives a wavelength near $10^{-10}\ \text{m}$. A 150 g ball at 30 m/s gives $10^{-34}\ \text{m}$. If your answer doesn't match this range, recheck.

5. Forgetting that the electron's mass is known

Some students don't know or forget the electron mass: $m_e = 9.1 \times 10^{-31}\ \text{kg}$. They might try to derive it or leave it out.

How to avoid: Memorise $m_e = 9.1 \times 10^{-31}\ \text{kg}$ as a standard constant for such problems.

Correct Solution

(a) Electron …