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CUET · PHYSICS · PYQ PAPER 2025

The shortest wavelength in the Balmer series of hydrogen atom would be (Given R = \( 1.1 \times 10^7 \, m^{-1} \))

  1. A 3636 \( \text{\AA} \)
  2. B 4529 \( \text{\AA} \)
  3. C 2433 \( \text{\AA} \)
  4. D 3454 \( \text{\AA} \)
Verified Solution

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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