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Q.

The ionization energy of the ground state hydrogen atom is $2.18 \times 10^{- 18} J$ . The energy of an electron in its second orbit would be

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a

$- 1.09 \times 10^{- 18} J$

b

$- 5.45 \times 10^{- 19} J$

c

$- 4.36 \times 10^{- 18} J$

d

$- 2.18 \times 10^{- 18} J$

answer is D.

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Detailed Solution

$E_{n} = - 2.18 \times 10^{- 18} \frac{Z^{2}}{n^{2}} J / a t m$

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