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Question

Chemistry Question on Structure of atom

The energy of an electron in second Bohr orbit of hydrogen atom is :

A

−5.44×10−19eV-5.44 \times 10^{-19} eV

B

−5.44×10−19 cal - 5.44 \times 10^{-19} \, cal

C

−5.44×10−19 kJ - 5.44 \times 10^{-19} \, kJ

D

−5.44×10−19 J - 5.44 \times 10^{-19} \, J

Answer

−5.44×10−19 J - 5.44 \times 10^{-19} \, J

Explanation

Solution

The energy of an electron in second Bohr orbit of hydrogen atom is
E=−13.6ev×(1n2)E =-13.6 ev \times\left(\frac{1}{ n ^{2}}\right)
n=2n =2
=−13.6×(122)=-13.6 \times\left(\frac{1}{2^{2}}\right)
=−13.64=−3.4ev=-\frac{13.6}{4}=-3.4 ev
Now we know, 1eV=1.6×10−19J1 eV =1.6 \times 10^{-19} J
∴E=−3.4×1.6×10−19\therefore E=-3.4 \times 1.6 \times 10^{-19}
=−5.44×10−19 J=-5.44 \times 10^{-19}\, J