Question
The species having identical radii according to the Bohr's theory are: A. H (first orbit) B. He^+ (first orbit) C. He^+ (Second orbit) D. Li^ 2+ (first orbit) E. Be^ 3+ (Second orbit) Choose the correct answer from the options given below:
JEE Main
Chemistry
Structure of Atom
2026
JEE Main 2026 (04 April Shift 2)
JEE Main Chemistry Question (2026) — Solution
Options
- A. A and C Only
- B. A and E Only
- C. B and E Only
- D. C and D Only
Answer
B. A and E Only
Step-by-step solution
According to Bohr's theory, the radius of the n-th orbit of a hydrogen-like species is given by r_n = 0.529 n^2 Z . For H (first orbit), n = 1, Z = 1 r = 0.529 1^2 1 = 0.529 . For He^+ (first orbit), n = 1, Z = 2 r = 0.529 1^2 2 = 0.2645 . For He^+ (second orbit), n = 2, Z = 2 r = 0.529 2^2 2 = 1.058 . For Li^ 2+ (first orbit), n = 1, Z = 3 r = 0.529 1^2 3 = 0.176 . For Be^ 3+ (second orbit), n = 2, Z = 4 r = 0.529 2^2 4 = 0.529 . The radii of H (first orbit) and Be^ 3+ (second orbit) are identical. Therefore, species A and E have identical radii. Answer: A and E Only
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