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The ratio of the energy required to remove an electron from the first three Bohr’s orbits of hydrogen is

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1 : 4 : 9

36 : 9 : 4

3 : 2 : 1

9 : 4 : 1

answer is C.

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Detailed Solution

$E_{n} = \frac{- 1312}{x^{2}} = \frac{x}{x^{2}} E_{1} = + \frac{x}{1^{2}} E_{2} = \frac{x}{2^{2}} E_{3} = \frac{x}{3^{2}} \frac{1}{1} : \frac{1}{4} : \frac{1}{9} = 36 : 9 : 4$

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