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

(D) $9.84 \times 10^{5} \mathrm{Jmol}^{-1}$

(A) $8.51 \times 10^{5} \mathrm{Jmol}^{-1}$

(B) $6.56 \times 10^{5} \mathrm{Jmol}^{-1}$

(C) $7.56 \times 10^{5} \mathrm{Jmol}^{-1}$

JEE Main · 2008 · 27 Apr 2008 · Offline · Atomic Structure

The ionization enthalpy of hydrogen atom is

1.312 \times 1 0^{6} Jmol^{- 1}

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

n = 1

n = 2

Answer: (D)

The ionisation enthalpy is the energy to remove the electron from the ground state, so the ground-state energy is

- 1.312 \times 1 0^{6} J mo l^{- 1}

and the

n = 2

level sits at one quarter of that depth,

- \frac{1}{4} \times 1.312 \times 1 0^{6} = - 3.28 \times 1 0^{5} J mo l^{- 1}

. Exciting the electron from

n = 1

n = 2

costs the difference,

1.312 \times 1 0^{6} - 3.28 \times 1 0^{5} = 9.84 \times 1 0^{5} J mo l^{- 1}

JEE Main · 2008 · 27 Apr 2008 · Offline · Single correct option