Energy of an electron is given by
\(E = -2.178 \times 10^{-18}\,\mathrm{J}\left(\frac{Z^2}{n^2}\right)\)
. Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be
(h = \(6.62 \times 10^{-34}\) Js and c = \(3.0 \times 10^{8}\) ms-1)
\(E = -2.178 \times 10^{-18}\,\mathrm{J}\left(\frac{Z^2}{n^2}\right)\)
. Wavelength of light required to excite an electron in an hydrogen atom from level n = 1 to n = 2 will be
(h = \(6.62 \times 10^{-34}\) Js and c = \(3.0 \times 10^{8}\) ms-1)
Show Hint
- \(2.816 \times 10^{-7}\) m
- \(6.500 \times 10^{-7}\) m
- \(8.500 \times 10^{-7}\) m
- \(1.214 \times 10^{-7}\) m
The Correct Option is D
Solution and Explanation
\(E_1 = -2.178 \times 10^{-18}\) J and \(E_2 = -2.178 \times 10^{-18}/4 = -0.5445 \times 10^{-18}\) J.
\(\Delta E = E_2 - E_1 = 2.178 \times 10^{-18}\left(1 - \frac{1}{4}\right) = 1.6335 \times 10^{-18}\) J.
\(\lambda = \frac{hc}{\Delta E} = \frac{6.62 \times 10^{-34} \times 3.0 \times 10^{8}}{1.6335 \times 10^{-18}} = \frac{1.986 \times 10^{-25}}{1.6335 \times 10^{-18}}\)
\(\lambda \approx 1.216 \times 10^{-7}\) m, closest to \(1.214 \times 10^{-7}\) m.
Answer: (4) \(1.214 \times 10^{-7}\) m
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