Question

Calculate the energy required to excite one litre of hydrogen gas at $1$ atm and $298K$ to the first excited state of atomic hydrogen. The energy for the dissociation of $H-H$ bond is $436kJ$ $mo{l}^{-I}$. Also calculate the minimum frequency of photon to break this bond.

• A. $E=107kJ$
  $v=4.5\times {10}^{13}se{c}^{-1}$
• B. $E=80.36kJ$
  $v=4.5\times {10}^{13}se{c}^{-1}$
• C. $E=5\times {10}^{23}eV$
  $v=10.93\times {10}^{14}se{c}^{-1}$
• D. $E=98.19kJ$
  $v=10.93\times {10}^{14}se{c}^{-1}$

Answer 1

Answer: (C), (D)

$E=5\times {10}^{23}eV$
$v=10.93\times {10}^{14}se{c}^{-1}$
$E=98.19kJ$

$v=10.93\times {10}^{14}se{c}^{-1}$
Mole of $H_2$ present in one litre$=PV/RT$
$=\frac{1\times 1}{0.0821\times 298}=0.0409mol$
Thus, energy needed to break $H-H$ bonds in $0.0409$ mole of $H_2$=$17.83kJ$
Also, energy needed to excite one $H-$ atom from $1st$ to $2nd$ energy level,
$\Delta E=E_2-E_1=\frac{E_1}{4}-E_1$ $(E_1=-13.6eV)$
$=13.6(1-\frac{1}{4})eV=10.2eV$
$=10.2\times 1.6\times {10}^{-19}J/atom$
One atom absorbs one quantum of energy
$\therefore$ Energy needed to excite $0.0409\times 2\times 6.02\times {10}^{23}$ atoms of $H$
$=10.2\times 1.6\times {10}^{-19}\times 0.0409\times 2\times 6.02\times {10}^{23}J$
$=80.36KJ$
Thus, total energy needed$=17.83+80.36=98.19KJ$
Energy required to break $(H-H)$ bond $=\frac{436\times {10}^3}{6.023\times {10}^{23}}$joule
$\because E=hv$
$\therefore \frac{436\times {10}^3}{6.023\times {10}^{23}}=6.626\times {10}^{-34}\times v$
$\therefore v=10.93\times {10}^{14}se{c}^{-1}$ or $Hz$.
Hence options C & D are correct.