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Question
Energy and radius of first Bohr orbit of He+ and Li2+ are:
[Given RH = −2.18 × 10−18 J, a0 = 52.9 pm]
Options
En (Li2+) = −19.62 × 10−18 J;
rn (Li2+) = 17.6 pm
En (He+) = −8.72 × 10−18 J;
rn (He+) = 26.4 pm
En (Li2+) = −8.72 × 10−18 J;
rn (Li2+) = 26.4 pm
En (He+) = −19.62 × 10−18 J;
rn (He+) = 17.6 pm
En (Li2+) = −19.62 × 10−16 J;
rn (Li2+) = 17.6 pm
En (He+) = −8.72 × 10−16 J;
rn (He+) = 26.4 pm
En (Li2+) = −8.72 × 10−16 J;
rn (Li2+) = 17.6 pm
En (He+) = −19.62 × 10−16 J;
rn (He+) = 17.6 pm
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Solution
En (Li2+) = −19.62 × 10−18 J;
rn (Li2+) = 17.6 pm
En (He+) = −8.72 × 10−18 J;
rn (He+) = 26.4 pm
Explanation:
Given: RH = −2.18 × 10−18 J
a0 = 52.9 pm
n = 1 (first orbit)
Z is the atomic number of He = 2
Z is the atomic number of Li = 3
Formula: `E_n = R_H (Z^2/n^2)`
`r_n = a_0 (n^2/Z)`
For He+ (Z = 2)
E1 = −2.18 × 10−18 (22/12)
= −2.18 × 10−18 (4/1)
= −2.18 × 10−18 × 4
= −8.72 × 10−18 J
`r_1 = 52.9 (1^2/2)`
= `52.9 (1/2)`
= 26.45 pm
For Li2+ (Z = 3)
E1 = `−2.18xx 10^(−18) (3^2/1^2)`
= `−2.18 xx 10^(−18) (9/1)`
= −2.18 × 10−18 × 9
= −19.62 × 10−18 J
`r_1 = 52.9 (1^2/3)`
= `52.9 (1/3)`
= 17.63 pm
