Question

An element undergoes a reaction as shown:
$X+2e^{-}\to X^{2-}$, energy released $=30.87$ eV/atom. If the energy released, is used to dissociate $4$ gms of $H_2$ molecules, equally into $H^{+}$ and $H^{\ast }$, where $H^{\ast }$ is excited state of H atoms where the electron travels in orbit whose circumference equal to four times ts de Broglie's wavelength. Determine the least moles of $X$ that would be required.
Given: I.E. of $H=13.6$ eV/atom, bond energy of $H_2=4.526$ eV/ molecule.

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is B $2$

Here each molecules of $H_2$

1) dissociates (require dissociate enthalpy)

2) $H^{+}$ (ionization enthalphy)

3) $H^{+}$ (excitation of H)

here H is 4 th state $\therefore$ Energy $=\frac{13.5}{n^2}=\frac{13.6}{16}$

$\therefore$ energy supplied $=13.6-\frac{13.6}{16}=\left(\frac{15}{16}\right)13.6$

$\therefore$ we have total $4H_2$ molecules & which means

$2Na{H}_2$ molecules.

$\therefore$ energy required

$\Rightarrow 2\times 6.022\times {10}^{23}$ (B.D.E+15.+Excitation energy)

$\Rightarrow 2\times 6.22\times {10}^{23}(4.576+13.6+(\frac{13}{16}\left)13.6\right)$

$\Rightarrow 2\times (2\times 6.022\times {10}^{23}\times 30.376)en/mole$

This energy is supplied by x atom

Energy released per atom = 30.37 ev/atom

so atoms req.

$\Rightarrow 2\times 6.022\times {10}^{23}\times 30.37ev/mole$

$n=2$

Ans (b)