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Question Number 14682 by tawa tawa last updated on 03/Jun/17

Answered by Tinkutara last updated on 03/Jun/17

$$\mathrm{K}.\mathrm{E}.=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0})$$$$=6.626\times {10}^{-34}\times 3\times {10}^{17}(\frac{1}{436}-\frac{1}{650})$$$$\approx 1.501\times {10}^{-19}\mathrm{J}$$

Commented by tawa tawa last updated on 03/Jun/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

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