Question

The ionization energy of hydrogen atom is $13.6$ eV. The energy required to excite the electron in a hydrogen atom from the ground state to the first excited state is:

• A. $1.69\times {10}^{-18}$ J
• B. $1.69\times {10}^{-23}$ J
• C. $1.69\times {10}^{23}$ J
• D. $1.69\times {10}^{25}$ J

Answer 1

Answer: (B)

$1.69\times {10}^{-23}$ J

$E=\frac{-13.6}{n^2}$

$=\frac{-13.6}{4}=-3.4$

$\Delta E=E_2-E_1$

$=-3.4-(-13.6)=10.2eV$

Hence the energy required for the excitation of electron pert atom.

$=\frac{10.2}{6.02\times {10}^{23}}=1.69\times {10}^{-23}J$