Two electrons, each moving with a velocity of ${10}^6 \mathrm{m}/{\mathrm{s}}^{-1}$ towards each other. The closest distance of approach between them is $\mathrm{y}\times {10}^{-10} \mathrm{m}.$ Find value of $\mathrm{y}.$

Using $\mathrm{s}, \mathrm{p}, \mathrm{d}$ notations, describe the orbital with the following quantum numbers.

 $\mathrm{n}=1, \mathrm{l}=0;$

How many electrons in an atom may have the following quantum numbers?

(a) $\mathrm{n}=4, {\mathrm{m}}_{\mathrm{s}}=-\frac{1}{2}$
(b) $\mathrm{n}=3, \mathrm{l}=0$

The quantum numbers of six electrons are given below. Arrange them in order of increasing energies. If any of these combination(s) has/have the same energy lists:

1. $\mathrm{n}=4, \mathrm{l}=2, {\mathrm{m}}_{\mathrm{l}}=–2, {\mathrm{m}}_{\mathrm{s}}=–1/2$
2. $\mathrm{n}=3, \mathrm{l}=2, {\mathrm{m}}_{\mathrm{l}}=1, {\mathrm{m}}_{\mathrm{s}}=+1/2$
3. $\mathrm{n}=4, \mathrm{l}=1, {\mathrm{m}}_{\mathrm{l}}=0, {\mathrm{m}}_{\mathrm{s}}=+1/2$
4. $\mathrm{n}=3, \mathrm{l}=2, {\mathrm{m}}_{\mathrm{l}}=–2, {\mathrm{m}}_{\mathrm{s}}=–1/2$
5. $\mathrm{n}=3, \mathrm{l}=1, {\mathrm{m}}_{\mathrm{l}}=–1, {\mathrm{m}}_{\mathrm{s}}=+1/2$
6. $\mathrm{n}=4, \mathrm{l}=1, {\mathrm{m}}_{\mathrm{l}}=0, {\mathrm{m}}_{\mathrm{s}}=+1/2$