JEE MAIN - Chemistry (2003 - No. 56)
The orbital angular momentum for an electron revolving in an orbit is given by $$\sqrt {l(l + 1)} {h \over {2\pi }}$$. This momentum for an s-electron will be given by
zero
$${h \over {2\pi }}$$
$$\sqrt 2 {h \over {2\pi }}$$
$$ + {1 \over 2}{h \over {2\pi }}$$
स्पष्टीकरण
For s-electron l = 0
$$\therefore $$ $$\sqrt {l(l + 1)} {h \over {2\pi }}$$
= $$\sqrt {0(0 + 1)} {h \over {2\pi }}$$
= $$\sqrt 0 {h \over {2\pi }}$$ = 0 (zero)
$$\therefore $$ $$\sqrt {l(l + 1)} {h \over {2\pi }}$$
= $$\sqrt {0(0 + 1)} {h \over {2\pi }}$$
= $$\sqrt 0 {h \over {2\pi }}$$ = 0 (zero)