Question

1. Calculate the orbital angular momentum for an electron having n=3,l=0
2. Calculate the maximum no. of electron having n+l=5 in an atom

Answer 1

Answer:

your answer is in the attachment

Answer 2

Answer:

1. The orbital angular momentum for an electron having n=3 and l=0 is 0.

The orbital angular momentum of an electron is given as follows:

Orbital angular momentum = $\sqrt{l(l+1)} \frac{h}{2\pi}$

Orbital angular momentum = $\sqrt{0(0+1)} \frac{h}{2\pi}$

Orbital angular momentum = 0

Therefore, the orbital angular momentum of an electron having n=3 and l=0 is zero.

2. The maximum number of electrons having n+l=5 is 18.

• The subshells having n+l = 5 are 5s, 4p, and 3d.
• The s subshell contains 2 electrons, p subshell contains 6 electrons, and d subshell contains 10 electrons at the maximum.
• Therefore, the maximum number of electrons will be 2+6+10 = 18.
• Hence, the maximum number of electrons with n+l=5 is 18.

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