Question

The ratio of the.energy required to remove electron
from second orbit and third orbit of He+ ion is
(1) 2:3
(2) 1:1
(3) 9:4
(4) 4:1-2​

Answer 1

Answer:

9:4

Explanation:

Energy is directly proportional to z square and inversely proportional to n square

• For 2nd orbit
• E=1/4
• For 3rd orbit
• E=1/9
• E2/3=9:4

Answer 2

Answer:

(3) 9:4

Explanation:

The expression for energy is:

$E_n=-Z^2\times 2.179\times 10^{-18}\times \frac{1}{n^2}\ Joules$

Where, n is the number of the energy level.

Z is the atomic number

For, He⁺ , Z = 2

For n = 2,  

$E_2=-2^2\times 2.179\times 10^{-18}\times \frac{1}{4}\ Joules$

For, n = 3,  

$E_3=-2^2\times 2.179\times 10^{-18}\times \frac{1}{9}\ Joules$

Ratio :

$\frac {E_2}{E_3}=\frac {-2^2\times 2.179\times 10^{-18}\times \frac{1}{4}}{-2^2\times 2.179\times 10^{-18}\times \frac{1}{9}}$

$\frac {E_2}{E_3}=\frac {\frac{1}{4}}{\frac{1}{9}}$

$\frac {E_2}{E_3}=\frac {9}{4}$