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Q.

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

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a

−1.09×10−18J

b

−5.45×10−19J

c

−4.36×10−18J

d

−2.18×10−18J

answer is D.

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Detailed Solution

En=−2.18×10−18Z2n2J/atm

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