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KCET · Physics · Atomic Physics

An electron in an excited state of \(\mathrm{Li}^{2+}\) ion has angular momentum \(\frac{3 h}{2 \pi}\). The de-Broglie wavelength of electron in this state is \(p \pi a_{0}\) (where, \(a_{0}=\) Bohr radius ). The value of \(p\) is

  1. A 3
  2. B 2
  3. C 1
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(B) 2

Step-by-step Solution

Detailed explanation

According to de-Broglie hypothesis,
\(=\frac{n h}{2 \pi}=\frac{3 h}{2 \pi}=m v r\)
\(\Rightarrow \quad n=3\)
As, wavelength, \(\lambda=\frac{h}{p}=\frac{h}{m v}=\frac{h r}{m v r}\)
\(=\frac{h r}{3 h}=\frac{2}{3} \pi r\)
For Li \({ }^{2+}\) atom, radius of orbit,
\(r=r_{0} \frac{n^{2}}{Z}=a_{0} \frac{3^{2}}{3}=3 a_{0}\)
\(\lambda=\frac{2}{3} \pi \times a_{0} \times 3=2 \pi a_{0}=p \pi a_{0} \text { (given) }\)
\(\quad p=2\)