Question

According to Bohr’s theory the energy required for an electron in the Li²⁺ ion to be emitted from n = 2 state is (given that the ground state ionization energy of hydrogen atom is 13.6 eV)
(a) 61.2 eV
(b) 13.6 eV
(c) 30.6 eV
(d) 10.2 eV

Answer 1

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Asked on December 27, 2019 by

Vibhusita Sisangia

The ionization energy of a hydrogen atom is 13.6 eV. The energy of the ground level in doubly ionized lithium is:

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ANSWER

Bohr's model is used to solve this question,

According to this E=−13.6×

n

2

Z

2

So, in this the E is equal to energy of electron in n

th

shell of atom whose atomic number is Z

.

Now, same amount of energy with opposite sign must be given to electron in order to remove it. I.E.=−13.6×Z

2

eV (As n=1)

Since for Lithium, Z=3

Therefore, you get I.E.=−122.4eV

Answer 2

13.6x2~30

answer -option c

this is the answer