Question

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

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

Answer 1

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

Given,
The ionization energy enthalpy of hydrogen atom $=1.312\times {10}^6\frac{Z^2}{n^2}J/mol$

Again,
Ionization energy (IE) = $-E_1$
$\text{Energy},E_n=-1.312\times {10}^6\frac{Z^2}{n^2}J/mol$
Hence,
The energy required to excite an electron in an atom of hydrogen from n = 1 to n = 2 is the energy differenece between the two energy levels, which is given by
$\Delta E=E_f-E_i$

$\therefore \Delta E=E_2-E_1$
$=(-1.312\times {10}^6\frac{1}{2^2})-(-1.312\times {10}^6\frac{1}{1^2})$
$=-1.312\times {10}^6\times (-\frac{3}{4})$
$=0.984\times {10}^{6}Jmo{l}^{-1}$
$=9.84\times {10}^{5}Jmo{l}^{-1}$