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Question: The ionization energy of the ground state hydrogen atom is \(2.18 \times 10^{- 18}J.\) The energy of...

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

A

−1.09×10−18J- 1.09 \times 10^{- 18}J

B

−2.18×10−18J- 2.18 \times 10^{- 18}J

C

−4.36×10−18J- 4.36 \times 10^{- 18}J

D

−5.45×10−19J- 5.45 \times 10^{- 19}J

Answer

−5.45×10−19J- 5.45 \times 10^{- 19}J

Explanation

Solution

Energy of electron in first Bohr's orbit of H-atom

E=−2.18×10−18n2JE = \frac{- 2.18 \times 10^{- 18}}{n^{2}}J (∵(\becauseionization energy of H =

2.18×10−18J2.18 \times 10^{- 18}J)

E2=−2.18×10−1822J=−5.45×10−19JE_{2} = \frac{- 2.18 \times 10^{- 18}}{2^{2}}J = - 5.45 \times 10^{- 19}J