The wavelength of a neutron with a translatory kinetic energy equal
to KT at 300 K is :-

Options

A. 178 pm
B. 200 pm
C. 17.8 pm
D. 20.0 pm

Explanation

To find the wavelength of a neutron with a translatory kinetic energy equal to \( kT \) at 300 K, we can follow these steps:

### Step 1: Understand the relationship between kinetic energy and wavelength
The de Broglie wavelength (\( \lambda \)) of a particle can be calculated using 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 kinetic energy
The momentum \( p \) of a particle can also be expressed in terms of its kinetic energy (\( KE \)):
\[ KE = \frac{p^2}{2m} \implies p = \sqrt{2m \cdot KE} \]
where \( m \) is the mass of the particle.

### Step 3: Substitute kinetic energy
Given that the translatory kinetic energy is equal to \( kT \), we can substitute this into the momentum equation:
\[ p = \sqrt{2m \cdot kT} \]

### Step 4: Substitute momentum into the wavelength formula
Now, substituting \( p \) into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m \cdot kT}} \]

### Step 5: Plug in known values
1. **Planck's constant \( h \)**: \( 6.626 \times 10^{-34} \, \text{J s} \)
2. **Mass of a neutron \( m \)**: \( 1.67 \times 10^{-27} \, \text{kg} \)
3. **Boltzmann's constant \( k \)**: \( 1.38 \times 10^{-23} \, \text{J/K} \)
4. **Temperature \( T \)**: \( 300 \, \text{K} \)

Now substituting these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (1.67 \times 10^{-27}) \times (1.38 \times 10^{-23}) \times 300}} \]

### Step 6: Calculate the denominator
First, calculate the term inside the square root:
\[ 2 \times (1.67 \times 10^{-27}) \times (1.38 \times 10^{-23}) \times 300 \]
Calculating this gives:
\[ = 2 \times 1.67 \times 1.38 \times 300 \times 10^{-50} \approx 2.079 \times 10^{-50} \]

### Step 7: Calculate the square root
Now, take the square root:
\[ \sqrt{2.079 \times 10^{-50}} \approx 4.56 \times 10^{-25} \]

### Step 8: Calculate the wavelength
Now substitute this back into the wavelength equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{4.56 \times 10^{-25}} \approx 1.45 \times 10^{-9} \, \text{m} \]
Converting to picometers:
\[ \lambda \approx 145 \, \text{pm} \]

### Final Answer
The wavelength of a neutron with a translatory kinetic energy equal to \( kT \) at 300 K is approximately **145 picometers**.