Question

Calculate the energy emitted when electrons of 1.0 gram atom of hydrogen undergo transition giving the spectral line of lowest wave energy in the visible region of its atomic spectrum.

• A. $n_2=3$ to $n_1=2;E=182.8KJ$
• B. $n_2=2$ to $n_1=1;E=155.8KJ$
• C. $n_2=3$ to $n_1=1;E=180.8KJ$
• D. $n_2=4$ to $n_1=2;E=182.5KJ$

Answer 1

The correct option is A $n_2=3$ to $n_1=2;E=182.8KJ$

For transition of lowest wave energy of Balmer series (visible), the transition is $n=3\to m=1$.
Therefore,
$\Delta E=13.6eV\times (1/m^2-1/n^2)=1.89eV=182kJmo{l}^{-1}$.