If the de Broglie wavelength of the electron in `n^(th)` Bohr orbit in a hydrogenic atom is equal to `1.5pia_(0)(a_(0)` is bohr radius), then the value of `n//z` is :

Options

A. `0.40`
B. `1.50`
C. `1.0`
D. `0.75`

Explanation

To solve the problem, we need to use the de Broglie wavelength concept and the Bohr model of the hydrogen atom. Here’s a step-by-step solution:

### Step 1: Understand the given information
We are given that the de Broglie wavelength (λ) of the electron in the nth Bohr orbit of a hydrogenic atom is equal to \(1.5 \pi a_0\), where \(a_0\) is the Bohr radius.

### Step 2: Use the de Broglie wavelength formula
The de Broglie wavelength of an electron is given by the formula:
\[ \lambda = \frac{h}{p} \]
where \(h\) is Planck's constant and \(p\) is the momentum of the electron. However, in the context of the Bohr model, we can relate it to the orbit radius.

### Step 3: Relate the de Broglie wavelength to the Bohr model
In the Bohr model, the circumference of the nth orbit is given by:
\[ 2 \pi r_n = n \lambda \]
where \(r_n\) is the radius of the nth orbit. The radius of the nth orbit in a hydrogenic atom is given by:
\[ r_n = \frac{n^2 a_0}{z} \]
where \(z\) is the atomic number.

### Step 4: Substitute the radius into the circumference equation
Substituting \(r_n\) into the circumference equation, we get:
\[ 2 \pi \left(\frac{n^2 a_0}{z}\right) = n \lambda \]

### Step 5: Substitute the value of λ
Now, substituting \(\lambda = 1.5 \pi a_0\):
\[ 2 \pi \left(\frac{n^2 a_0}{z}\right) = n \left(1.5 \pi a_0\right) \]

### Step 6: Simplify the equation
We can cancel \(a_0\) and \(\pi\) from both sides:
\[ 2 \left(\frac{n^2}{z}\right) = 1.5 n \]

### Step 7: Rearranging the equation
Now, rearranging gives:
\[ \frac{2n^2}{z} = 1.5n \]

### Step 8: Solve for \(\frac{n}{z}\)
Dividing both sides by \(n\) (assuming \(n \neq 0\)):
\[ \frac{2n}{z} = 1.5 \]
Now, solving for \(\frac{n}{z}\):
\[ \frac{n}{z} = \frac{1.5}{2} = 0.75 \]

### Final Answer
Thus, the value of \(\frac{n}{z}\) is \(0.75\).

---