Assume that the de Broglie wave associated with an electron can from a standing wave between the atom arrange in a one dimensional array with nodes at each of the atomic sites. It is found that one such standing wave if the distance d between the atoms of the array is `2 A^0` A similar standing wave is again formed if d is increased to `2.5Å` . Find the energy of the electrons in electron volts and the least value of d for which the standing wave type described above can from .

Explanation

To solve the problem step by step, we will follow the reasoning provided in the video transcript, while ensuring clarity and completeness in each step.

### Step 1: Understand the Standing Wave Condition
The standing wave condition for an electron in a one-dimensional array of atoms states that the distance between the nodes of the standing wave should be equal to an integer multiple of the wavelength (λ). The distance between the atomic sites is given as \(d\).

### Step 2: Set Up the Equations
For the first case where \(d = 2 \, \text{Å}\):
\[ d = n \frac{\lambda}{2} \quad \text{(1)} \]
For the second case where \(d = 2.5 \, \text{Å}\):
\[ d = (n + 1) \frac{\lambda}{2} \quad \text{(2)} \]

### Step 3: Substitute the Values
Substituting the values of \(d\) into the equations:
1. From equation (1):
\[ 2 = n \frac{\lambda}{2} \implies \lambda = \frac{4}{n} \quad \text{(3)} \]
2. From equation (2):
\[ 2.5 = (n + 1) \frac{\lambda}{2} \implies \lambda = \frac{5}{n + 1} \quad \text{(4)} \]

### Step 4: Equate the Two Expressions for λ
From equations (3) and (4), we can set them equal to each other:
\[ \frac{4}{n} = \frac{5}{n + 1} \]

### Step 5: Cross Multiply and Solve for n
Cross multiplying gives:
\[ 4(n + 1) = 5n \]
Expanding and rearranging:
\[ 4n + 4 = 5n \implies n = 4 \]

### Step 6: Find the Wavelength (λ)
Substituting \(n = 4\) back into equation (3):
\[ \lambda = \frac{4}{4} = 1 \, \text{Å} \]

### Step 7: Calculate the Energy of the Electron
The energy \(E\) of the electron can be calculated using the de Broglie wavelength formula:
\[ E = \frac{h^2}{2m\lambda^2} \]
Where:
- \(h = 6.63 \times 10^{-34} \, \text{Js}\)
- \(m = 9.1 \times 10^{-31} \, \text{kg}\)
- \(\lambda = 1 \times 10^{-10} \, \text{m}\)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34})^2}{2 \times (9.1 \times 10^{-31}) \times (1 \times 10^{-10})^2} \]

### Step 8: Perform the Calculation
Calculating the above expression:
1. Calculate \(h^2\):
\[ h^2 = 4.39 \times 10^{-67} \, \text{Js}^2 \]
2. Calculate \(2m\lambda^2\):
\[ 2m\lambda^2 = 2 \times (9.1 \times 10^{-31}) \times (1 \times 10^{-10})^2 = 1.82 \times 10^{-50} \, \text{kg m}^2 \]
3. Therefore,
\[ E = \frac{4.39 \times 10^{-67}}{1.82 \times 10^{-50}} \approx 241 \, \text{J} \]
4. Convert to electron volts (1 eV = \(1.6 \times 10^{-19} \, \text{J}\)):
\[ E \approx \frac{241}{1.6 \times 10^{-19}} \approx 151 \, \text{eV} \]

### Step 9: Find the Least Value of d
The least value of \(d\) occurs when \(n = 1\):
\[ d_{\text{min}} = \frac{\lambda}{2} = \frac{1 \, \text{Å}}{2} = 0.5 \, \text{Å} \]

### Final Answers
- The energy of the electrons is approximately **151 eV**.
- The least value of \(d\) for which the standing wave can form is **0.5 Å**.