An electron is an excited state of `Li^(2 + )`ion has angular momentum `3h//2 pi ` . The de Broglie wavelength of the electron in this state is `p pi a_(0)` (where `a_(0) ` is the Bohr radius ) The value of p is

Options

A. p = 2
B. p=4
C. p=1
D. p=3

Answer

To solve the problem step by step, we will analyze the given information and apply the relevant physics concepts.

### Step 1: Understand the given data
We know that the angular momentum \( L \) of the electron in the excited state of the \( \text{Li}^{2+} \) ion is given as:
\[ L = \frac{3h}{2\pi} \]
where \( h \) is Planck's constant.

### Step 2: Relate angular momentum to quantum number
The angular momentum of an electron in a hydrogen-like atom is quantized and can be expressed as:
\[ L = n \frac{h}{2\pi} \]
where \( n \) is the principal quantum number. From the given angular momentum, we can set:
\[ n \frac{h}{2\pi} = \frac{3h}{2\pi} \]
By comparing both sides, we find:
\[ n = 3 \]

### Step 3: Write the expression for de Broglie wavelength
The de Broglie wavelength \( \lambda \) of a particle is given by:
\[ \lambda = \frac{h}{p} \]
where \( p \) is the momentum of the electron. The momentum can also be expressed as:
\[ p = mv \]
where \( m \) is the mass of the electron and \( v \) is its velocity.

### Step 4: Relate de Broglie wavelength to radius and quantum number
From the expression for angular momentum, we have:
\[ L = mvR = n \frac{h}{2\pi} \]
We can rearrange this to find \( mv \):
\[ mv = \frac{n h}{2\pi R} \]

### Step 5: Substitute into the de Broglie wavelength equation
Substituting \( mv \) into the de Broglie wavelength equation:
\[ \lambda = \frac{h}{mv} = \frac{h}{\frac{n h}{2\pi R}} = \frac{2\pi R}{n} \]

### Step 6: Find the radius of the orbit
The radius \( R_n \) of the \( n \)-th orbit in a hydrogen-like atom is given by:
\[ R_n = \frac{n^2 a_0}{Z} \]
where \( a_0 \) is the Bohr radius and \( Z \) is the atomic number. For \( \text{Li}^{2+} \), \( Z = 3 \).

### Step 7: Substitute the radius into the wavelength equation
Now substituting \( R_n \) into the wavelength equation:
\[ \lambda = \frac{2\pi \left(\frac{n^2 a_0}{Z}\right)}{n} = \frac{2\pi n a_0}{Z} \]
Substituting \( n = 3 \) and \( Z = 3 \):
\[ \lambda = \frac{2\pi (3) a_0}{3} = 2\pi a_0 \]

### Step 8: Compare with the given wavelength expression
We are given that:
\[ \lambda = p \pi a_0 \]
Comparing both expressions:
\[ 2\pi a_0 = p \pi a_0 \]
Dividing both sides by \( \pi a_0 \):
\[ 2 = p \]

### Conclusion
Thus, the value of \( p \) is:
\[ \boxed{2} \]