Question Number 14682 by tawa tawa last updated on 03/Jun/17
Answered by Tinkutara last updated on 03/Jun/17
$$\mathrm{K}.\mathrm{E}.\:=\:{hc}\left(\frac{\mathrm{1}}{\lambda}\:−\:\frac{\mathrm{1}}{\lambda_{\mathrm{0}} }\right) \\ $$$$=\:\mathrm{6}.\mathrm{626}×\mathrm{10}^{−\mathrm{34}} ×\mathrm{3}×\mathrm{10}^{\mathrm{17}} \left(\frac{\mathrm{1}}{\mathrm{436}}\:−\:\frac{\mathrm{1}}{\mathrm{650}}\right) \\ $$$$\approx\:\mathrm{1}.\mathrm{501}\:×\:\mathrm{10}^{−\mathrm{19}} \:\mathrm{J} \\ $$
Commented by tawa tawa last updated on 03/Jun/17
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$