The angular momentum of the electron in first excited energy state of hydrogen atom is

Question

The angular momentum of the electron in first excited energy state of hydrogen atom is (A)  (h/2π )  (B)  (h/π )  (C)  √2(2+l)(h/2π )  (D) none of these

Tardigrade Admin · Accepted Answer

The angular momentum of an electron =mvr=(nh/2π ) For the first excited state n=2 then, mvr=(2h/2π ) =(h/π )

Q. The angular momentum of the electron in first excited energy state of hydrogen atom is

$\frac{h}{2 π}$

$\frac{h}{π}$

$2 (2 + l) \frac{h}{2 π}$

none of these

The angular momentum of an electron
$= m v r = \frac{nh}{2 π}$
For the first excited state $n = 2$
then, $m v r = \frac{2 h}{2 π}$
$= \frac{h}{π}$

Q. The angular momentum of the electron in first excited energy state of hydrogen atom is

2πh​

πh​

2(2+l)​2πh​

none of these

The angular momentum of an electron =mvr=2πnh​ For the first excited state n=2 then, mvr=2π2h​ =πh​

$\frac{h}{2 π}$

$\frac{h}{π}$

$2 (2 + l) \frac{h}{2 π}$

The angular momentum of an electron
$= m v r = \frac{nh}{2 π}$
For the first excited state $n = 2$
then, $m v r = \frac{2 h}{2 π}$
$= \frac{h}{π}$