Q23 / 105 · JEE Main 2008 · Chemistry

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 $n=1$ to $n=2$ is

  • A
    $8.51 \times 10^5 \mathrm{Jmol}^{-1}$
  • B
    $6.56 \times 10^5 \mathrm{Jmol}^{-1}$
  • C
    $7.56 \times 10^5 \mathrm{Jmol}^{-1}$
  • Correct answer
    D
    $9.84 \times 10^5 \mathrm{Jmol}^{-1}$

Solution

$$ \begin{aligned} & \Delta \mathrm{E}=\mathrm{E}_2-\mathrm{E}_1=-\frac{1.312 \times 10^6}{2^2}-\left(-\frac{1.312 \times 10^6}{1}\right) \\ & =9.84 \times 10^5 \mathrm{~J} \mathrm{~mol}^{-1} \end{aligned} $$

Hence, the correct answer is option: 'D'.

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