Question

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

• A. $-1.09\times {10}^{-18}J$
• B. $-2.18\times {10}^{-18}J$
• C. $-4.36\times {10}^{-18}J$
• D. $-5.45\times {10}^{-19}J$

Answer 1

Answer: (D)

$-5.45\times {10}^{-19}J$

The ionization energy is given to be $=2.18\times {10}^{-18}J$, which equals $13.6eV$. Therefore, the energy of electron in the first level is $-13.6eV$. We can thus use the formula, $E=-13.6\times Z^2/n^2$ to calculate energy of electron in the second orbit.

$E_2=\frac{-2.18\times {10}^{-18}}{2^2}=-5.45\times {10}^{-19}J$