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The ionization energy of hydrogen atom is 13.6 eV. Following Bohr's theory, the energy corresponding to a transition between the 3^rd and the 4^th orbit is

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3.40 eV

1.51 eV

0.85 eV

0.66 eV

answer is D.

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Detailed Solution

$n = 3 z = 1 n = 4 z = 1$

$E_{3} = \frac{- 13.6}{3^{2}} E_{4} = \frac{- 13.6}{4^{2}}$

$= \frac{- 13.6}{9} = \frac{- 13.6}{16}$

$= - 1.51 = - 0.85$ $E_{4} - E_{3} = - 0.85 - (- 1.51)$

$= - 0.85 + 1.51 = 0.66 eV$

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