Based on the equation ΔE=−2.0×10−18J1n22−1n12, the wavelength of light that must be absorbed to excite hydrogen electron from level n=1 to level n=2 will be (Given: h=6.625×10−34Js,c=3×108ms−1 )

# Based on the equation $\mathrm{\Delta E}=-\left(2.0×{10}^{-18}\mathrm{J}\right)\left(\frac{1}{{\mathrm{n}}_{2}^{2}}-\frac{1}{{\mathrm{n}}_{1}^{2}}\right)$, the wavelength of light that must be absorbed to excite hydrogen electron from level $\mathrm{n}=1$ to level $\mathrm{n}=2$ will be (Given:

1. A

$1.325×{10}^{-7}\mathrm{m}$

2. B

$1.325×{10}^{-10}\mathrm{m}$

3. C

$2.650×{10}^{-7}\mathrm{m}$

4. D

$5.300×{10}^{-10}\mathrm{m}$

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### Solution:

$\mathrm{\lambda }=\frac{\mathrm{hc}}{\mathrm{\Delta E}}=\frac{\left(6.625×{10}^{-34}\mathrm{Js}\right)\left(3×{10}^{8}{\mathrm{ms}}^{-1}\right)}{\left(1.5×{10}^{-18}\mathrm{J}\right)}=1.325×{10}^{-7}\mathrm{m}$

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