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According to the Bohr Theory, which of the following transitions in the hydrogen atom will give rise to the least energetic photon?

Option: 1

n = 6 to n = 1


Option: 2

n = 5 to n = 4


Option: 3

n = 5 to n = 3


Option: 4

n = 6 to n = 5


Answers (1)

best_answer

According to Bohr's theory,

\Delta \mathbf{E} \propto\left[1 / n_{1}^{2}-1 / n_{2}^{2}\right]

now, let's check all option

\\(a) \ n=6\ to \ n=1 \\ \left[1 / 1^{2}-1 / 6^{2}\right] = [1-1 / 36]=35 / 36\ \\ \\(b) \ n=5\ to \ n=4 \\ \left[1 / 4^{2}-1 / 5^{2}\right] = [1 / 16-1 / 25]=9 / 400\ \\ \\ (c) \ n=5\ to \ n=3 \\ \left[1 / 3^{2}-1 / 5^{2}\right] = [1 / 9-1 / 25 ]=16 / 225\ \\ \\ (d) \ n=6\ to \ n=5 \\ \left[15^{2}-1 / 6^{2}\right]=1 / 25-1 / 36=1 / 900

\\\text{Here least value of} \ \left[1 / n_{1}^{2}-1 / n_{2}^{2}\right]\ \text{for option} \(d)\ \\\text{Hence, answer is option} \ (d) \ n=6\ to \ n=5\\

Posted by

Rakesh

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