Question

$1.8g$ hydrogen atoms are excited to radiation. The study of spectra indicates that $27\%$ of the atoms are in 3${}^{rd}$ energy level and $15\%$ of atoms in 2${}^{nd}$ energy level and the rest in the ground state. $IP$ of $H$ is $21.7\times {10}^{-12}erg$. Calculate total energy in kJ evolved when all the atoms return to ground state.

• A. 758KJ
• B. 832KJ
• C. 1058KJ
• D. None of these
  

Answer 1

Answer: (B)

832KJ

$\because 1g$ $H$ contains $N$ atoms $H$

$\therefore 1.8g$ $H$ contains $N\times 1.8$ atoms of $H$

$=6.023\times {10}^{23}\times 1.8=10.84\times {10}^{23}$ atoms

$\left(a\right)$ $\therefore$ No.of atoms in 3${}^{rd}$ shell

$=10.8\times {10}^{23}\times \frac{27}{100}=292.68\times {10}^{21}$ atoms
No. of atoms in II shell

$=10.84\times {10}^{23}\times \frac{15}{100}=162.6\times {10}^{21}$ atoms
No.of atoms in I shell

$=10.84\times {10}^{23}\times \frac{58}{100}=628.72\times {10}^{21}$ atoms
$\left(b\right)$ When all the excited atoms return to the normal state, the energy-related = Energy released during de-excitation from 2${}^{nd}$ to 1${}^{st}$ $+$ Energy released during de-excitation from 3${}^{rd}$ to 1${}^{st}$

$=[E_2-E_1]\times 162.6\times {10}^{21}+[E_3-E_1]\times 292.68\times {10}^{21}$

$=[-(13.6/4)-(-13.6\left)\right]\times 162.6\times {10}^{21}+[-(13.6/9)-(-13.6\left)\right]\times 292.68\times {10}^{21}$

$=1.658\times {10}^{24}+3.538\times {10}^{24}$

$=5.196\times {10}^{24}\times 1.602\times {10}^{-19}J$

$=5.196\times {10}^{24}\times 1.602\times {10}^{-19}\times {10}^{-3}KJ$

$=832.40KJ$

Hence, the correct option is $B$