CUET · PHYSICS · PYQ PAPER 2025
The shortest wavelength in the Balmer series of hydrogen atom would be (Given R = \( 1.1 \times 10^7 \, m^{-1} \))
- A 3636 \( \text{\AA} \)
- B 4529 \( \text{\AA} \)
- C 2433 \( \text{\AA} \)
- D 3454 \( \text{\AA} \)
Answer & Solution
Correct Answer
(A) 3636 \( \text{\AA} \)
Step-by-step Solution
Detailed explanation
\( \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) \( \frac{1}{\lambda} = 1.1 \times 10^7 \, m^{-1} \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) \) \( \frac{1}{\lambda} = 1.1 \times 10^7 \, m^{-1} \left( \frac{1}{4} - 0 \right) \)…
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