Question

The ionisation energy of H - atom is 1.312 x ${10}^6\parallel mol$. calculate the energy require to excite an e - in an H - atom from the ground state to the 1st excited state? $\left(M^{22}\right)N_A=6.02\times 1023$

Answer 1

Ionisation energy for $4-$ atom

$=\frac{1.32\times {10}^6}{6.02\times {10}^{23}}$

$\therefore$ Energy required to excite an $e^{-}$ in an H atom from the ground state (n=1) to the $1^{st}$ excited state (n=2)

$=\frac{1.312\times {10}^6}{6.02\times {10}^{23}}(\frac{1}{1^2}-\frac{1}{(2{)}^2})$

$=\frac{1.312\times {10}^6}{6.02\times {10}^{23}}\times \frac{3}{4}$

$=0.1635\times {10}^{-17}$

$=16.35\times {10}^{-19}J$