Question

Comprehension-3
Electron present in single electron species jumps from energy level 3 to 1. Emitted photons when passed through
a sample containing excited He+ ion causes further excitation to some higher energy level.
z?
(Given E - 13,62)
17.
(D) 4
Determine atomic number of single electron species.
(A) 1
(B) 2
(C)3
Determine Principal quantum number of initial excited level of He.
(A) 1
(B) 2
(C)4
18.
(D) 6
19.
Determine Principal quantum number of higher energy level of He'.
(A) 2
(B) 3
(C) 4
(D) 6​

Answer 1

Answer:

2

3

4

6

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Answer 2

Given info : Electron present in single electron species jumps from energy level 3 to 1. Emitted photons when passed through a sample containing excited He+ ion causes further excitation to some higher energy level.

To find : (A) Determine atomic number of single electron species.

(B) Determine Principal quantum number of initial excited level of He⁺.

(C) Determine Principal quantum number of higher energy level of He⁺.

Solution : using formula, -13.6 Z²[1/n₁² - 1/n₂²]

Case 1 : for single electron species, atomic no Z , n₁ = 1 and n₂ = 3

So, E = -13.6 × Z²[1 - 1/3²] = -13.6 × Z × 8/9 eV .......(1)

case 2 : for helium ion, He⁺ Z = 2,

E = -13.6 × 2² [1/n₁² - 1/n₂²] eV ......(2)

from equations (1) and (2) we get,

-13.6 × Z² × 8/9 eV = -13.6 × 2² [1/n₁² - 1/n₂²]

⇒Z²8/9 = 4[1/n₁² - 1/n₂²]

⇒2Z²/9 = [1/n₁² - 1/n₂²]

If we assume, Z = 1

Then, 2/9 = [1/n₁² - 1/n₂²]

it is possible only when n₁ = 2 and n₂ = 6 [ here 1/2² - 1/6² = 8/36 = 2/9 ]

so atomic no of single electron species is 1.

n = 2 of helium ion of initial excited state and n = 6 of higher excited state.