-->

Wave Optics

Question

If an electron has orbital angular momentum quantum number l =7 then it will have an orbital angular momentum equal to

$\sqrt{56} (\frac{h}{2 π})$

$57 (\frac{h}{2 π})$

$42 (\frac{h}{2 π})$

$7 (\frac{h}{2 π})$

Solution

$\sqrt{56} (\frac{h}{2 π})$

$L = \sqrt{l (l + 1)} . \frac{h}{2 π} = \sqrt{7 \times 8} \frac{h}{2 π} = \sqrt{56} \frac{h}{2 π}$