Question

The ionization enthalpy of hydrogen atom is $1.312\times {10}^{6}J$ mol${}^{-1}$. The energy required to excite the electron in the atom from $n=1$ to $n=2$ is :

• A. $7.56\times {10}^{5}Jmo{1}^{-1}$
• B. $9.84\times {10}^{5}Jmo{1}^{-1}$
• C. $8.51$ $\times {10}^{5}Jmo{1}^{-1}$
• D. $6.56\times {10}^{5}Jmo{1}^{-1}$

Answer 1

The correct option is B $9.84\times {10}^{5}Jmo{1}^{-1}$

(a): The ionization of H-atom is the energy absorbed when the electron in

first shell $\left(E_1\right)$ to infinity (i.e.,$E_{\infty }$)

$I.E=E_{\infty }-E_1$

$1.312\times {10}^6=0-E_1$

$E_1=-1.312\times {10}^{6}Jmo{l}^{-1}$

$E_2=-\frac{1.312\times {10}^6}{(2{)}^2}=-\frac{1.312\times {10}^6}{4}$

$\therefore$ Energy required when an electron makes

transition from $n=1$ to $n=2$

$\Delta E=E_2-E_1=-\frac{1.312\times {10}^6}{4}-(-1.312\times {10}^6)$

$\Delta E=\frac{-1.312\times {10}^6+5.248\times {10}^6}{4}=0.984\times {10}^6$

$\Delta E=9.84\times {10}^{5}Jmo{l}^{-1}$