The orbital angular momentum of an electron in 2s-orbital is
- $+\frac{1}{2}.\frac{h}{2 \pi}$
- zero
- $\frac{h}{2 \pi}$
- $\sqrt2.\frac{h}{2 \pi}$
The Correct Option is B
Solution and Explanation
$ \, \, \, \, \, \, \, \, L=\sqrt{l(l +1)} \frac{h}{2 \pi}=0 \, \, \, \, \, $ for 2s-electrons
$\because \, \, $ For s-orbital, l= 0.
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Given below are two statements. One is labelled as Assertion (A) and the other is labelled as Reason (R).
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Concepts Used:
Momentum
It can be defined as "mass in motion." All objects have mass; so if an object is moving, then it is called as momentum.
the momentum of an object is the product of mass of the object and the velocity of the object.
Momentum = mass • velocity
The above equation can be rewritten as
p = m • v
where m is the mass and v is the velocity.
Momentum is a vector quantity and the direction of the of the vector is the same as the direction that an object.