Question

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

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

Answer 1

Answer: (A)

The ionisation of $H$ - atom is the energy absorbed when the electron in an atom gets excited from first shell $\left(E_1\right)$ to infinity $($ i.e. $,E_{\infty })$.

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

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

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

$E_2=-\frac{1.312\times 10^6}{(2{)}^2}=-\frac{1.312\times 10^6}{4}Jmo{l}^{-1}$

Energy of electron in second orbit $(n=2)$

$\therefore$ Energy required when an electron makes transition from $n=1$ to $n=2$

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

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

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