The de - Broglie wavelength of a neutron in thermal equilibrium with heavy water at a temperature `T ("kelvin")` and `mass m`, is

Options

A. `(h)/(sqrt((3 mkT))`
B. `( 2h)/(sqrt((3 mkT))`
C. `(2h)/(sqrt(( mkT))`
D. `(h)/(sqrt(( mkT))`

Explanation

To find the de Broglie wavelength of a neutron in thermal equilibrium with heavy water at temperature \( T \) (in Kelvin) and mass \( m \), we can follow these steps:

### Step 1: Determine the thermal kinetic energy of the neutron.
The thermal kinetic energy \( KE \) of a particle in thermal equilibrium is given by the formula:
\[ KE = \frac{3}{2} k T \]
where \( k \) is the Boltzmann constant and \( T \) is the temperature in Kelvin.

### Step 2: Relate kinetic energy to momentum.
The kinetic energy can also be expressed in terms of momentum \( p \) as:
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the neutron. Setting the two expressions for kinetic energy equal gives us:
\[ \frac{3}{2} k T = \frac{p^2}{2m} \]

### Step 3: Solve for momentum \( p \).
Rearranging the equation to solve for \( p \):
\[ p^2 = 3mkT \]
Taking the square root of both sides, we find:
\[ p = \sqrt{3mkT} \]

### Step 4: Use 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. Substituting our expression for momentum into this formula gives:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

### Final Answer:
Thus, the de Broglie wavelength of a neutron in thermal equilibrium with heavy water at temperature \( T \) is:
\[ \lambda = \frac{h}{\sqrt{3mkT}} \]

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