Which of the following has the largest de Broglie wavelength (all have equal velocity)?

Options

A. (a) `CO_(2)` molecule
B. (b) `NH_(3)` molecule
C. (c) Electron
D. (d) Proton

Answer

To determine which of the given options has the largest de Broglie wavelength when all have equal velocity, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (λ) is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle.

### Step 2: Relate momentum to mass and velocity
Momentum (p) can be expressed as:
\[ p = mv \]
where \(m\) is the mass of the particle and \(v\) is its velocity. Therefore, we can rewrite the de Broglie wavelength as:
\[ \lambda = \frac{h}{mv} \]

### Step 3: Analyze the relationship between wavelength and mass
From the equation \(\lambda = \frac{h}{mv}\), we can see that the wavelength is inversely proportional to the mass when the velocity is constant:
\[ \lambda \propto \frac{1}{m} \]
This means that the particle with the smallest mass will have the largest de Broglie wavelength.

### Step 4: Calculate the mass of each option
1. **CO2 molecule**:
- Mass of Carbon (C) = 12 amu
- Mass of Oxygen (O) = 16 amu
- Total mass of CO2 = 12 + (2 × 16) = 44 amu

2. **NH3 molecule**:
- Mass of Nitrogen (N) = 14 amu
- Mass of Hydrogen (H) = 1 amu
- Total mass of NH3 = 14 + (3 × 1) = 17 amu

3. **Electron**:
- Mass of electron = \(9.1 \times 10^{-31}\) kg (or \(9.1 \times 10^{-28}\) grams)

### Step 5: Convert amu to grams for comparison
1 amu = \(1.66 \times 10^{-24}\) grams.
- Mass of CO2 in grams = \(44 \times 1.66 \times 10^{-24} \approx 7.304 \times 10^{-23}\) grams
- Mass of NH3 in grams = \(17 \times 1.66 \times 10^{-24} \approx 2.822 \times 10^{-23}\) grams

### Step 6: Compare the masses
- Mass of CO2 = \(7.304 \times 10^{-23}\) grams
- Mass of NH3 = \(2.822 \times 10^{-23}\) grams
- Mass of electron = \(9.1 \times 10^{-28}\) grams

### Step 7: Identify the smallest mass
The electron has the smallest mass compared to CO2 and NH3.

### Conclusion
Since the electron has the smallest mass, it will have the largest de Broglie wavelength. Therefore, the answer is:
**The electron has the largest de Broglie wavelength.**