Question

An electron in an excited state of Li²⁺ ion has angular momentum \(\frac{3h}{2\pi}\). The de Broglie wavelength of electron in this state is Pπa₀ (where a₀ = Boh r radius). The value of P is

• A. 3
• B. 2
• C. 1
• D. 4

Answer 1

The Correct Option is B

Given the angular momentum quantization condition:

\[ L = \frac{nh}{2\pi} = \frac{3h}{2\pi} \quad \text{(where } n = 3\text{)} \]

The de Broglie wavelength is calculated as:

\[ \lambda = \frac{h}{mv} = \frac{hr}{mvr} = \frac{hr}{\frac{3h}{2\pi}} = \frac{2}{3}\pi r \]

For a Li²⁺ ion, the radius of the electron orbit is given by:

\[ r = r_0\frac{n^2}{z} \]

\[ r = a_0 \times \frac{3^2}{3} \quad \text{(where } a_0 \text{ is the Bohr radius)} \]

Substituting the radius into the wavelength equation:

\[ \lambda = \frac{2}{3}\pi \times a_0 \times \frac{3^2}{3} = 2\pi a_0 \]

Comparing with the given form \(\lambda = P\pi a_0\), we find:

\[ P = 2 \]

Answer 2

1. Determine Principal Quantum Number (n)

Angular momentum = \(\frac{nh}{2\pi}\) = \(\frac{3h}{2\pi}\)
Therefore, n = 3

2. Calculate Radius of the Orbit (r_n)

Radius \(r_n = \frac{a_0 n^2}{Z}\)
For Li²⁺, Z = 3 and n = 3
\(r_3 = \frac{a_0 (3)^2}{3} = 3a_0\)

3. Calculate de Broglie Wavelength (λ)

Circumference = nλ => \(2\pi r_n = n\lambda\)
\(\lambda = \frac{2\pi r_n}{n} = \frac{2\pi (3a_0)}{3} = 2\pi a_0\)

4. Find the Value of P

Given de Broglie wavelength = Pπa₀
Comparing with calculated wavelength: Pπa₀ = \(2\pi a_0\)
Therefore, P = 2

Final Answer: The final answer is $2$