The energy and radius of the first Bohr orbit are given by the following formulas:
The energy of the nth orbit for a hydrogen-like atom:
\(E_n = - \frac{R_H}{n^2} Z^2\)
Where \( Z \) is the atomic number of the ion.
The radius of the nth orbit:
\(r_n = \frac{a_0}{Z} \cdot n\)
For Li2+ (Z = 3) and He+ (Z = 2), we calculate for \( n = 1 \) (the first Bohr orbit).
For Li2+ (Z = 3):
\(E_n (\text{Li}^{2+}) = - \frac{2.18 \times 10^{-18}}{1^2} \cdot 3^2 = -19.62 \times 10^{-18} \, \text{J}\)
\(r_n (\text{Li}^{2+}) = \frac{52.9 \, \text{pm}}{3} = 17.6 \, \text{pm}\)
For He+ (Z = 2):
\(E_n (\text{He}^{+}) = - \frac{2.18 \times 10^{-18}}{1^2} \cdot 2^2 = -8.72 \times 10^{-18} \, \text{J}\)
\(r_n (\text{He}^{+}) = \frac{52.9 \, \text{pm}}{2} = 26.4 \, \text{pm}\)
Thus, the correct values are:
Therefore, the correct answer is (1).
An element has two isotopes having atomic masses 10 and 15 u, respectively. If the percent abundance of lighter isotopes is 80%, then the average atomic mass of the element is: