Question

The hydrogen like species (He+) is in spherically symmetric state S1 with two

radial nodes. Upon absorbing light, the ion undergoes transition to state S2 which

has two radial nodes and has energy equal to (1/4) times the ground state energy

of H-atom

Q1 The state S1 is

A) 1s B) 2s C) 4p D) 3s

Q2 The sum of principal quantum number and azimuthal quantum number of higher

state S2-

A) 3 B) 4 C) 5 D) 7

Q3 The number of angular node in higher state S2 will be

A) 0 B) 1 C) 2 D) 3 ​

Answer 1

Answer:

the hydrogen like species Li2+ in a spherically symmetric state SI with one radial node and it's energy it's equal to the ground state energy of the hydrogen atom

Answer 2

Answer:

1. The state $S_1$ is 3s.

2. The sum of the principal quantum number and azimuthal quantum number of higher state S2 is 4+1 = 5

3. The number of angular nodes in the state $S_2$ is equal to azimuthal quantum number which is 1.  

Explanation:

$S_1$ symmetrical implies l=0

No. of nodes = (n-l-1) = 2

                      =n-0-1=2

                      n=3

With n=3 l=0 the orbital is 3s.

$S_2$ has two radial nodes

energy of $S_2$ = $\frac{1}{4}$ Energy of ground state H atom

energy of $S_2$ = $\frac{1}{4} \frac{-Z^2R}{n^2}$

$\frac{-Z^2R}{n^2}$ = $\frac{1}{4} \frac{-1^2R}{1^2}$

$\frac{-2^2R}{n^2} = \frac{-R}{4}\\ 16 = n^2\\4 =n$

$S_2$ has two radial nodes

n-l-1 = 2

4-l-1=2

3-l=2

l= 1

So The sum of the principal quantum number and azimuthal quantum number of higher state S2 is 4+1 = 5

The number of angular nodes in the state $S_2$ is equal to azimuthal The number of angular nodes that is 1.