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NEET Chemistry
QUESTION #7381
Question 1
In a hydrogen atom, electron transitions occur from \(n=5\) to \(n=2\), emitting a photon of wavelength 434 nm. What will be the wavelength of the photon emitted when the electron transitions from \(n=4\) to \(n=2\)?
Correct Answer Explanation
Using Rydberg formula: \(\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{n_1^2} - \dfrac{1}{n_2^2}\right)\)
For \(n=4 \to n=2\) (H\(\beta\) line of Balmer series): \(\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = R_H \times \dfrac{3}{16}\)
This gives \(\lambda \approx 486\) nm (blue-green, visible Balmer series H\(\beta\) line). The 434 nm line (\(n=5\to2\)) is H\(\gamma\). These are well-known spectral lines of hydrogen.
For \(n=4 \to n=2\) (H\(\beta\) line of Balmer series): \(\dfrac{1}{\lambda} = R_H\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = R_H \times \dfrac{3}{16}\)
This gives \(\lambda \approx 486\) nm (blue-green, visible Balmer series H\(\beta\) line). The 434 nm line (\(n=5\to2\)) is H\(\gamma\). These are well-known spectral lines of hydrogen.
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