Question

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

• A. $1.825\times {10}^{-4}$ J
• B. $18.25\times {10}^4$ J
• C. $3.03\times {10}^4$ J
• D. $3.03\times {10}^{19}$ J

Answer 1

The correct option is B $18.25\times {10}^4$ J

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