Question

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

Answer 1

$E_2=\frac{-1.32\times {10}^6\times (1{)}^2}{2^2}$

$⟹E_2=-3.28\times {10}^5$ $Jmo{l}^{-1}$

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

$\Delta E=E_2-E_1$

$⟹\Delta E=-3.28\times {10}^5-(-1.32\times {10}^6)$

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