Question

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

A
n=5 to n=3
B
n=6 to n=1
C
n=5 to n=4
D
n=6 to n=5

Solution

## The correct option is C $$n = 6$$ to $$n = 5$$$$E=-13.6\cfrac { { z }^{ 2 } }{ { n }^{ 2 } } eV$$. Thus $$E\propto \cfrac { 1 }{ { n }^{ 2 } }$$.For the transition $$n_2 \rightarrow n_1$$ we have,$$\Delta E \propto (\dfrac{1}{n_1^2}- \dfrac{1}{n_2^2})$$ If we observe this relationship closely, we can observe more higher value of n, more less the value of E. Thus least energetic photon is observed in transition of $$n=6$$ to $$n=5$$. Thus answer is option D.Chemistry

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