The energies of an electron in first orbit `He ^(+)` and in third orbit of ` Li^(2+) ` in J are respectively

Options

A. `- 872xx10^(-18), - 2.18 xx10^(-18)`
B. `- 872xx10^(-18), - 1.96 xx10 ^(-17)`
C. `-1.96 xx10^(-17), - 2.18 xx 10^(-18)`
D. `- 8.72xx10^(-17), - 1.96 xx 10^(-17)`

Answer

A

Explanation

Key Idea Use relation:
` E_(n) = - 2.18 xx10 ^(-18) .(Z^(2))/(n^(n)) J//ion`
where,
` E_(n) `= energy of an electron in nth orbit
Z = atmic number , n = ortit number
For helium ion ` (He^(+))`
Z = 2 , n = 1
` therefore E_(n) = - 2.18 xx 10^(-18) xx (2^(2))/(1^(2))`
` = -8.72 xx 10^(-18) J`
For lithium ion ` (Li^(2+))`
Z = 3 , n = 3
` therefore E_(n) = - 2.18 xx 10^(-18) xx (3^(2))/(3^(2))`
` = - 218 xx 10^(-18) J`
Hence, option (a) is the correct answer .