Question
The ionization enthalpy of hydrogen atom is $$1.312 \times {10^6}\,J\,mo{l^{ - 1}}.$$ The energy required to excite the electron in the atom from $$n = 1$$ to $$n = 2$$ is
A.
$$8.51 \times {10^5}\,J\,mo{l^{ - 1}}$$
B.
$$6.56 \times {10^5}\,J\,mo{l^{ - 1}}$$
C.
$$7.56 \times {10^5}\,J\,mo{l^{ - 1}}$$
D.
$$9.84 \times {10^5}\,J\,mo{l^{ - 1}}$$
Answer :
$$9.84 \times {10^5}\,J\,mo{l^{ - 1}}$$
Solution :
$$\left( {\Delta E} \right),$$ The energy required to excite an electron in an atom of hydrogen from $$n = 1$$ to $$n=2$$ is $${\Delta E}$$ (difference in energy $${E_1}$$ and $${E_2}$$
Values of $${E_2}$$ and $${E_1}$$ are,
$$\eqalign{
& {E_1} = - 1.312 \times {10^6}J\,mo{l^{ - 1}} \cr
& {E_2} = \frac{{ - 1.312 \times {{10}^6} \times {{\left( 1 \right)}^2}}}{{{{\left( 2 \right)}^2}}} = - 3.28 \times {10^5}J\,mo{l^{ - 1}} \cr
& \Delta E\,{\text{is}}\,{\text{given}}\,{\text{by}}\,{\text{the}}\,{\text{relation,}} \cr
& \therefore \,\,\Delta E = {E_2} - {E_1} = \left[ { - 3.28 \times {{10}^5}} \right] - \left[ { - 1.312 \times {{10}^6}} \right]\,J\,mo{l^{ - 1}} \cr
& = \left( { - 3.28 \times {{10}^5} + 1.312 \times {{10}^6}} \right)\,J\,mo{l^{ - 1}} \cr
& = 9.84 \times {10^5}\,J\,mo{l^{ - 1}} \cr} $$
Thus the correct answer is (D)