The ionisation enthalpy of hydrogen atom is $1.312 \times {10^6}{\rm{Jmo}}{{\rm{l}}^{ - 1}}$. The energy required to excite the electron in the atom from ${n_1} = 1\;{\rm{to }}\;{n_2} = 2$ is

$9.84 \times {10^5}{\rm{\;J\;mo}}{{\rm{l}}^{ - 1}}$

$8.51 \times {10^{5{\rm{\;}}}}{\rm{J\;mo}}{{\rm{l}}^{ - 1}}$

$6.56 \times {10^5}{\rm{\;J\;mo}}{{\rm{l}}^{ - 1}}$

$7.56 \times {10^5}{\rm{\;J\;mo}}{{\rm{l}}^{ - 1}}$

JEE CHEMISTRY MCQ: STRUCTURE OF ATOM

1 Hard

1 of 1

Application +4 | -1

The ionisation enthalpy of hydrogen atom is

1.312 \times {10^6}\mathrm{Jmo}{\mathrm{l}^{ - 1}}

. The energy required to excite the electron in the atom from

{n_1} = 1\;\mathrm{to }\;{n_2} = 2

View answer and explanation

Correct option: D

9.84 \times {10^5}\mathrm{\;J\;mo}{\mathrm{l}^{ - 1}}

The correct answer is

9.84 \times {10^5}\mathrm{\;J\;mo}{\mathrm{l}^{ - 1}}

because it is calculated by finding the difference in energy levels between n=1 and n=2 using the ionization enthalpy.