Question

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

• A. -5.44 $\times {10}^{-19}$ eV
• B. -5.44 $\times {10}^{-19}$ cal
• C. -5.44 $\times {10}^{-19}$ kJ
• D. -5.44 $\times {10}^{-19}$ J

Answer 1

Answer: (D)

-5.44 $\times {10}^{-19}$ J

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

$E=-13.6ev\times \left(\frac{1}{n^2}\right)n=2=-13.6\times \left(\frac{1}{2^2}\right)=-\frac{13.6}{4}=-3.4ev$

Now we know, $1eV=1.6\times {10}^{-19}J$

$\therefore E=-3.4\times 1.6\times {10}^{-19}=-5.44\times {10}^{-19}J$