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Based on the equation:
$$
\Delta \mathrm{E}=-2.0 \times 10^{-18} \mathrm{~J}\left(\frac{1}{\mathrm{n}_2^2}-\frac{1}{\mathrm{n}_1^2}\right)
$$
the wavelength of the light that must be absorbed to excite hydrogen electron from level $n=1$ to level $\mathrm{n}=2$ will be: $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}, \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}\right)$
Options:
$$
\Delta \mathrm{E}=-2.0 \times 10^{-18} \mathrm{~J}\left(\frac{1}{\mathrm{n}_2^2}-\frac{1}{\mathrm{n}_1^2}\right)
$$
the wavelength of the light that must be absorbed to excite hydrogen electron from level $n=1$ to level $\mathrm{n}=2$ will be: $\left(\mathrm{h}=6.625 \times 10^{-34} \mathrm{Js}, \mathrm{C}=3 \times 10^8 \mathrm{~ms}^{-1}\right)$
Solution:
1161 Upvotes
Verified Answer
The correct answer is:
$1.325 \times 10^{-7} \mathrm{~m}$
$1.325 \times 10^{-7} \mathrm{~m}$
$$
\begin{aligned}
&\Delta E=-2.0 \times 10^{-18} \times\left(\frac{1}{2^2}-\frac{1}{1^2}\right) \\
&=-2.0 \times 10^{-18} \times \frac{-3}{4} \\
&=1.5 \times 10^{-18} \\
&\Delta E=\frac{h c}{\lambda} \\
&\lambda=\frac{h c}{\Delta E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.5 \times 10^{-18}} \\
&=1.325 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$
\begin{aligned}
&\Delta E=-2.0 \times 10^{-18} \times\left(\frac{1}{2^2}-\frac{1}{1^2}\right) \\
&=-2.0 \times 10^{-18} \times \frac{-3}{4} \\
&=1.5 \times 10^{-18} \\
&\Delta E=\frac{h c}{\lambda} \\
&\lambda=\frac{h c}{\Delta E}=\frac{6.6 \times 10^{-34} \times 3 \times 10^8}{1.5 \times 10^{-18}} \\
&=1.325 \times 10^{-7} \mathrm{~m}
\end{aligned}
$$
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