Energy and radius of the ground-state Bohr orbit of He+ and Li^2+ are: [Given R_H = 2.18 × 10-18 J, a_0 = 52.9 pm]

Correct answer: (C) E_n(Li^2+) = -19.62 × 10-18 J, r_n(Li^2+) = 17.6 pm; E_n(He+) = -8.72 × 10-18 J, r_n(He+) = 26.4 pm. Answer: (C) E_n(Li^2+) = -19.62 × 10-18 J, r_n(Li^2+) = 17.6 pm; E_n(He+) = -8.72 × 10-18 J, r_n(He+) = 26.4 pm Solution: For the ground state (n=1): E_n = -R_H Z2n2 and r_n = a_0 n2Z. For Li^2+ (Z=3): E_n = -2.18 10-18× 9 = -19.62 10-18 J, r_n = 52.93 = 17.6 pm. For He+ (Z=2): E_n = -2.18 10-18× 4 = -8.72 10-18 J, r_n = 52.92 = 26.4 pm.

NEET 2025 · ChemistryPrevious Year Question

Energy and radius of the ground-state Bohr orbit of $He X^{+}$ and $Li X^{2 +}$ are: [Given $R_{H} = 2.18 \times 1 0^{- 18}$ J, $a_{0} = 52.9$ pm]

Answer: (C) $E_{n} (Li X^{2 +}) = - 19.62 \times 1 0^{- 18}$ J, $r_{n} (Li X^{2 +}) = 17.6$ pm; $E_{n} (He X^{+}) = - 8.72 \times 1 0^{- 18}$ J, $r_{n} (He X^{+}) = 26.4$ pm. \textbf{Answer:} (C) $E_{n} (Li X^{2 +}) = - 19.62 \times 1 0^{- 18}$ J, $r_{n} (Li X^{2 +}) = 17.6$ pm; $E_{n} (He X^{+}) = - 8.72 \times 1 0^{- 18}$ J, $r_{n} (He X^{+}) = 26.4$ pm \textbf{Solution:} For the ground state ( $n = 1$ ): $E_{n} = - R_{H} \frac{Z ^{2}}{n ^{2}}$ and $r_{n} = a_{0} \frac{n ^{2}}{Z}$ .

A. $E_{n} (Li X^{2 +}) = - 19.62 \times 1 0^{- 16}$ J, $r_{n} (Li X^{2 +}) = 17.6$ pm; $E_{n} (He X^{+}) = 8.72 \times 1 0^{- 16}$ J, $r_{n} (He X^{+}) = 26.4$ pm
B. $E_{n} (Li X^{2 +}) = - 8.72 \times 1 0^{- 16}$ J, $r_{n} (Li X^{2 +}) = 17.6$ pm; $E_{n} (He X^{+}) = - 19.62 \times 1 0^{- 16}$ J, $r_{n} (He X^{+}) = 17.6$ pm
C. $E_{n} (Li X^{2 +}) = - 19.62 \times 1 0^{- 18}$ J, $r_{n} (Li X^{2 +}) = 17.6$ pm; $E_{n} (He X^{+}) = - 8.72 \times 1 0^{- 18}$ J, $r_{n} (He X^{+}) = 26.4$ pm✓
D. $E_{n} (Li X^{2 +}) = - 8.72 \times 1 0^{- 18}$ J, $r_{n} (Li X^{2 +}) = 26.4$ pm; $E_{n} (He X^{+}) = - 19.62 \times 1 0^{- 18}$ J, $r_{n} (He X^{+}) = 17.6$ pm

Correct Answer

(C) $E_{n} (Li X^{2 +}) = - 19.62 \times 1 0^{- 18}$ J, $r_{n} (Li X^{2 +}) = 17.6$ pm; $E_{n} (He X^{+}) = - 8.72 \times 1 0^{- 18}$ J, $r_{n} (He X^{+}) = 26.4$ pm

Solution & Explanation

\textbf{Answer:} (C) $E_{n} (Li X^{2 +}) = - 19.62 \times 1 0^{- 18}$ J, $r_{n} (Li X^{2 +}) = 17.6$ pm; $E_{n} (He X^{+}) = - 8.72 \times 1 0^{- 18}$ J, $r_{n} (He X^{+}) = 26.4$ pm \textbf{Solution:} For the ground state ( $n = 1$ ): $E_{n} = - R_{H} \frac{Z ^{2}}{n ^{2}}$ and $r_{n} = a_{0} \frac{n ^{2}}{Z}$ . For $Li X^{2 +}$ ( $Z = 3$ ): $E_{n} = - 2.18 \times 1 0^{- 18} \times 9 = - 19.62 \times 1 0^{- 18}$ J, $r_{n} = \frac{52.9}{3} = 17.6$ pm. For $He X^{+}$ ( $Z = 2$ ): $E_{n} = - 2.18 \times 1 0^{- 18} \times 4 = - 8.72 \times 1 0^{- 18}$ J, $r_{n} = \frac{52.9}{2} = 26.4$ pm.

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Energy and radius of the ground-state Bohr orbit of HeX+ and LiX2+ are: [Given RH​=2.18×10−18 J, a0​=52.9 pm]

Solution & Explanation

More NEET PYQ solutions

Energy and radius of the ground-state Bohr orbit of $He X^{+}$ and $Li X^{2 +}$ are: [Given $R_{H} = 2.18 \times 1 0^{- 18}$ J, $a_{0} = 52.9$ pm]