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: (Avogadro's constant =6.022 × 10²³)

• A. 1.69 × 10^-20 J
• B. 1.69 × 10^-23 J
• C. 1.69 ×10 ²³ J
• D. 1.69 ×10 ²⁵ J

Answer 1

The correct option is B

1.69 × 10^-23 J

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

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

We know that energy required for excitation is $△$ E=$E_2-E_1$=-3.4-(-13.6)=10.2 eV

Therefore energy required for excitation of electron per atom =$\frac{10.2}{6.02\times {10}^{23}}$=$1.69\times {10}^{-23}$J