The de Broglie wavelength of neutrons in thermal equilibrium is (given `m_n=1.6xx10^(-27)`kg)`

Options

A. `(30.8)/(sqrtT)A`
B. `(3.08)/(sqrtT)A`
C. `(0.308)/(sqrtT)A`
D. `(0.0308)/(sqrtT)A`

Explanation

To find the de Broglie wavelength of neutrons in thermal equilibrium, we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) 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 kinetic energy
For a particle in thermal equilibrium, the kinetic energy (\( E \)) is given by:
\[ E = kT \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin. The momentum can be expressed in terms of kinetic energy as:
\[ p = \sqrt{2mE} = \sqrt{2m(kT)} \]
where \( m \) is the mass of the neutron.

### Step 3: Substitute momentum into the de Broglie wavelength formula
Substituting the expression for momentum into the de Broglie wavelength formula gives:
\[ \lambda = \frac{h}{\sqrt{2m(kT)}} \]

### Step 4: Substitute known values
Given:
- Mass of neutron, \( m_n = 1.6 \times 10^{-27} \) kg
- Planck's constant, \( h = 6.626 \times 10^{-34} \) J·s
- Boltzmann constant, \( k = 1.38 \times 10^{-23} \) J/K

Substituting these values into the equation:
\[ \lambda = \frac{6.626 \times 10^{-34}}{\sqrt{2 \times (1.6 \times 10^{-27}) \times (1.38 \times 10^{-23}) \times T}} \]

### Step 5: Simplify the expression
Calculating the denominator:
\[ \sqrt{2 \times (1.6 \times 10^{-27}) \times (1.38 \times 10^{-23})} = \sqrt{4.416 \times 10^{-50}} = 6.646 \times 10^{-25} \]
Thus, the expression for wavelength becomes:
\[ \lambda = \frac{6.626 \times 10^{-34}}{6.646 \times 10^{-25} \sqrt{T}} = \frac{6.626}{6.646} \times 10^{-9} \times \frac{1}{\sqrt{T}} \approx 0.996 \times 10^{-9} \frac{1}{\sqrt{T}} \]

### Step 6: Convert to appropriate units
To express the wavelength in Ångströms (1 Å = \( 10^{-10} \) m):
\[ \lambda \approx \frac{30.8}{\sqrt{T}} \text{ Å} \]

### Final Answer
The de Broglie wavelength of neutrons in thermal equilibrium is:
\[ \lambda \approx \frac{30.8}{\sqrt{T}} \text{ Å} \]