The wavelength of a neutron with energy 1eV is closest to 0

Options

A. `10^(-2) cm`
B. `10^(-4) cm `
C. `10^(-6) cm `
D. `10^(-8) cm`

Answer

To find the wavelength of a neutron with energy 1 eV, we can use the de Broglie wavelength formula and the relationship between energy and momentum. Here’s a step-by-step solution:

### Step 1: Understand the relationship between energy and wavelength
The de Broglie wavelength (λ) of a particle is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the particle. The momentum \(p\) can be expressed in terms of energy \(E\) as:
\[ p = \sqrt{2mE} \]
where \(m\) is the mass of the particle and \(E\) is its energy.

### Step 2: Substitute the expression for momentum into the wavelength formula
Substituting the expression for momentum into the wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2mE}} \]

### Step 3: Insert known values
- Planck's constant \(h = 6.626 \times 10^{-34} \, \text{J s}\)
- Mass of a neutron \(m \approx 1.675 \times 10^{-27} \, \text{kg}\)
- Energy \(E = 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J}\)

### Step 4: Calculate the wavelength
Now, substituting these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times 1.675 \times 10^{-27} \times 1.6 \times 10^{-19}}} \]

### Step 5: Calculate the denominator
First, calculate the term under the square root:
\[ 2 \times 1.675 \times 10^{-27} \times 1.6 \times 10^{-19} = 5.36 \times 10^{-46} \]
Now, take the square root:
\[ \sqrt{5.36 \times 10^{-46}} \approx 7.33 \times 10^{-23} \]

### Step 6: Calculate the wavelength
Now substitute back into the wavelength equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{7.33 \times 10^{-23}} \approx 9.05 \times 10^{-12} \, \text{m} \]

### Step 7: Convert to centimeters
To convert meters to centimeters, multiply by \(100\):
\[ \lambda \approx 9.05 \times 10^{-12} \, \text{m} \times 100 = 9.05 \times 10^{-10} \, \text{cm} \]

### Step 8: Final approximation
The closest value in scientific notation is:
\[ \lambda \approx 0.905 \times 10^{-9} \, \text{cm} \approx 0.9 \times 10^{-9} \, \text{cm} \]

### Conclusion
Thus, the wavelength of a neutron with energy 1 eV is closest to \(0.9 \times 10^{-9} \, \text{cm}\).