Q144 / 170 · Manipal (MET) 2015 · Physics

In wave mechanics, the angular momentum of an electron is given by

  • A
    $\sqrt{1 \frac{h}{2 \pi}}$
  • Correct answer
    B
    $\sqrt{l(l+1) \frac{h}{2 \pi}}$
  • C
    $\sqrt{\frac{H R \pi}{1(1+1)}}$
  • D
    $\sqrt{\frac{H-2 \pi}{l}}$

Solution

In wave mechanics, the magnitude of angular momentum of an electron is $L=\sqrt{l(l+1) \frac{h}{2 \pi}}$ where, $l$ can have the values $0,1,2 \ldots \ldots(n+1)$ $n$ is principal quantum number.

Hence, the correct answer is option: 'B'.

Relevant Questions