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Question

Arrange the following set of quantum numbers having the highest energy of an electron.

(p) n=4,l=1,m=+1,s=+1/2

(q) n=4,l=1,m=-1,s=-1/2

(r) n=3,l=1,m=-1,s=-1/2

(s) n=3,l=1,m=+1,s=-1/2


A

(q) > (r) > (p) > (s)

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B

(q) > (p) > (r) >(s)

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C

(s) > (p) > (r) > (q)

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D

(s) > (r) > (p) > (q)

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Solution

The correct option is B

(q) > (p) > (r) >(s)


The explanation for the correct option:

(B)

  1. The higher the value of n (principal quantum number), the higher the energy of the electron.
  2. When the value of n becomes equal to two sets of values then the higher the value of (n+l) , the higher the energy of the electron.

On the basis of point 1 we can say that the energy of an electron in case of (p) and (q) is greater than (r) and (s). The value of n for (p) and (q) is 4 and for (s) and (t) which is 3.

So, we can write energy order as (p), (q) > (r), (s)

Now, value of (n+l) for (p) is =4+1

n+l=5

Now, value of (n+l) for (q) is =4+2

n+l=6

We can write energy order as (q) > (p)

Now, value of (n+l) for (r) is =3+2

n+l=5

Now, value of (n+l) for (s) is =3+1

n+l=4

We can write energy order as (r) >(s)

Final energy order can be written as (q) > (p) > (r) > (s)

Explanation for incorrect option:

(A) The above detailed explanation shows that option(A) is incorrect.

(C) The above detailed explanation shows that option(C) is incorrect

(D) The above detailed explanation shows that option(D) is incorrect

Hence, option B is the correct answer.


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