The frequency of radiation released when an electron in a hydrogen atom drops from n = 4 to n = 1 is: (Given: ionisation energy of H = 2.18 times 10^ -18 J atom^ -1 , h = 6.625 times 10^ -34 Js)

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NEET Chemistry QUESTION #7348

Question 1

The frequency of radiation released when an electron in a hydrogen atom drops from \(n = 4\) to \(n = 1\) is: (Given: ionisation energy of H \(= 2.18 \times 10^{-18}\) J atom\(^{-1}\), \(h = 6.625 \times 10^{-34}\) Js)

\(1.03 \times 10^{15}\) s\(^{-1}\)
\(3.08 \times 10^{15}\) s\(^{-1}\)✔️
\(2.00 \times 10^{15}\) s\(^{-1}\)
\(1.54 \times 10^{15}\) s\(^{-1}\)

Correct Answer Explanation

Step 1: \(\Delta E = E_1 \left(1 - \dfrac{1}{n^2}\right) = 2.18 \times 10^{-18}\left(1 - \dfrac{1}{16}\right) = 2.18 \times 10^{-18} \times \dfrac{15}{16}\)

Step 2: \(\nu = \dfrac{\Delta E}{h} = \dfrac{2.04 \times 10^{-18}}{6.625 \times 10^{-34}} \approx 3.08 \times 10^{15}\) s\(^{-1}\)

This transition belongs to the Lyman series (UV region).

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