The de-Broglie wavelength of electron in a certain orbit 'n' of `Be^"3+"` ion is found to be 5.83 Å.Find the value of 'n'.Take `I^"st"` Bohr radius for H-atom =53 pm :

Answer

To find the value of 'n' for the de-Broglie wavelength of an electron in the `Be^3+` ion, we can follow these steps:

### Step 1: Understand the de-Broglie wavelength formula
The de-Broglie wavelength (\( \lambda \)) of an electron in a circular orbit is given by the relationship:
\[ 2\pi r = n\lambda \]
where \( r \) is the radius of the orbit, \( n \) is the principal quantum number, and \( \lambda \) is the de-Broglie wavelength.

### Step 2: Determine the expression for the radius
For a hydrogen-like atom, the radius of the orbit can be expressed as:
\[ r = a_0 \frac{n^2}{Z} \]
where \( a_0 \) is the first Bohr radius (53 pm or \( 53 \times 10^{-12} \) m) and \( Z \) is the atomic number. For beryllium (\( Be \)), the atomic number \( Z = 4 \).

### Step 3: Substitute the radius into the de-Broglie equation
Substituting the expression for \( r \) into the de-Broglie wavelength equation:
\[ 2\pi \left(a_0 \frac{n^2}{Z}\right) = n\lambda \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ n = \frac{Z \lambda}{2\pi a_0} \]

### Step 5: Substitute the known values
Now, substitute \( Z = 4 \), \( \lambda = 5.83 \times 10^{-10} \) m (since 1 Å = \( 10^{-10} \) m), and \( a_0 = 53 \times 10^{-12} \) m:
\[ n = \frac{4 \times (5.83 \times 10^{-10})}{2 \times 3.14 \times (53 \times 10^{-12})} \]

### Step 6: Calculate the denominator
Calculating the denominator:
\[ 2 \times 3.14 \times (53 \times 10^{-12}) \approx 3.34 \times 10^{-10} \text{ m} \]

### Step 7: Calculate 'n'
Now substituting back into the equation for 'n':
\[ n = \frac{4 \times 5.83 \times 10^{-10}}{3.34 \times 10^{-10}} \approx \frac{23.32 \times 10^{-10}}{3.34 \times 10^{-10}} \approx 7 \]

Thus, the value of \( n \) is approximately \( 7 \).

### Answer
The value of \( n \) is \( 7 \).

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