Question

Q.90
The ratio of the energy required to remove electron from
second orbit and third orbit of He ion is
Choose answer:
2:3
1:1
9:4
4:1​

Answer 1

Answer:

2:3

Explanation:

Answer 2

Answer:

9:4

Explanation:

The expression for energy is:

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

Where, n is the number of the energy level.

For n = 2,

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

For, n = 3,

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

Ratio :

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

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

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