Question

Calculate the energy emitted when electrons of $1.0g$ atom of hydrogen undergo transition giving the spectral line of lowest energy in the visible region of its atomic spectrum.
$(R_H=1.1\times {10}^7{m}^{-1};c=3\times {10}^{8}m{s}^{-1};h=6.62\times {10}^{-34}J-s)$.

Answer 1

The transition occurs like Balmer series as spectral line is observed in visible region.
Thus, the line of lowest energy will be observed when transition occurs from $3$rd orbit to $2$nd orbit, i.e., $n_1=2$ and $n_2=3$.
$\frac{1}{\lambda }=R[\frac{1}{2^2}-\frac{1}{3^2}]=\frac{5}{36}R$
$E=hv=h\cdot \frac{c}{\lambda }=6.62\times {10}^{-34}\times 3\times {10}^8\times \frac{5}{36}\times 1.1\times {10}^7$
$=3.03\times {10}^{-19}$ $J$ per atom
Energy corresponding to $1.0g$ atom of hydrogen
$=3.03\times {10}^{-19}\times$ Avogadro's number
$=3.03\times {10}^{-19}\times 6.023\times {10}^{23}J$
$=18.25\times {10}^{4}J$.