The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which
A neutron, and would have the same de Broglie wavelength.

Options

A. `3.81 xx 10^(-28) J`
B. `3.81 xx 10^(-25) J`
C. `4.81 xx 10^(-28)J`
D. `4.81 xx 10^(-30)J`

Answer

To find the kinetic energy of a neutron that has the same de Broglie wavelength as the light from the spectral emission line of sodium (589 nm), we can follow these steps:

### Step 1: Understand the de Broglie wavelength formula
The de Broglie wavelength (\( \lambda \)) is given by the formula:
\[ \lambda = \frac{h}{mv} \]
where:
- \( h \) is the Planck constant (\( 6.626 \times 10^{-34} \, \text{Js} \)),
- \( m \) is the mass of the particle (for a neutron, \( m \approx 1.675 \times 10^{-27} \, \text{kg} \)),
- \( v \) is the velocity of the particle.

### Step 2: Rearrange the formula to find kinetic energy
The kinetic energy (\( K \)) of a particle can be expressed in terms of momentum (\( p \)):
\[ K = \frac{p^2}{2m} \]
Since momentum \( p \) can also be expressed as:
\[ p = mv = \frac{h}{\lambda} \]
we can substitute this into the kinetic energy equation:
\[ K = \frac{\left(\frac{h}{\lambda}\right)^2}{2m} \]

### Step 3: Substitute known values
Now, we can substitute the known values into the equation:
- Convert the wavelength from nanometers to meters:
\[ \lambda = 589 \, \text{nm} = 589 \times 10^{-9} \, \text{m} \]
- Substitute \( h \), \( \lambda \), and \( m \):
\[ K = \frac{\left(6.626 \times 10^{-34}\right)^2}{2 \times 1.675 \times 10^{-27} \times 589 \times 10^{-9}} \]

### Step 4: Calculate \( K \)
1. Calculate \( h^2 \):
\[ h^2 = (6.626 \times 10^{-34})^2 = 4.39 \times 10^{-67} \, \text{J}^2\text{s}^2 \]
2. Calculate \( 2m \):
\[ 2m = 2 \times 1.675 \times 10^{-27} = 3.35 \times 10^{-27} \, \text{kg} \]
3. Calculate \( \lambda^2 \):
\[ \lambda^2 = (589 \times 10^{-9})^2 = 3.469 \times 10^{-13} \, \text{m}^2 \]
4. Substitute these values into the kinetic energy formula:
\[ K = \frac{4.39 \times 10^{-67}}{3.35 \times 10^{-27} \times 3.469 \times 10^{-13}} \]
5. Calculate the denominator:
\[ 3.35 \times 10^{-27} \times 3.469 \times 10^{-13} \approx 1.16 \times 10^{-39} \]
6. Finally, calculate \( K \):
\[ K = \frac{4.39 \times 10^{-67}}{1.16 \times 10^{-39}} \approx 3.78 \times 10^{-28} \, \text{J} \]

### Final Answer
The kinetic energy at which a neutron would have the same de Broglie wavelength as the sodium spectral line is approximately:
\[ K \approx 3.78 \times 10^{-28} \, \text{J} \]