JEE Mains · Chemistry · STD 11 - 2. structure of atom
The shortest wavelength of hydrogen atom in Lyman series is \(\lambda\). The longest wavelength in Balmer series of \(He ^{+}\)is
- A \(\frac{5}{9 \lambda}\)
- B \(\frac{9 \lambda}{5}\)
- C \(\frac{36 \lambda}{5}\)
- D \(\frac{5 \lambda}{9}\)
Answer & Solution
Correct Answer
(B) \(\frac{9 \lambda}{5}\)
Step-by-step Solution
Detailed explanation
For \(H : \frac{1}{\lambda}= R _{ H } \times 1^2\left(\frac{1}{1^2}-\frac{1}{\infty^2}\right)\) \(\frac{1}{\lambda_{ He ^{+}}}= R _{ H } \times 2^2 \times\left(\frac{1}{4}-\frac{1}{9}\right)\) From \((1)\) and \((2)\) \(\frac{\lambda_{ He ^{+}}}{\lambda}=\frac{9}{5}\)…
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