Atomic StructureHard
Question
A photon of 2.55 eV is emitted out by an electronic transition in hydrogen atom. The change in de Broglie wavelength of the electron is
Options
A.3.32 Å
B.4.98 Å
C.6.64 Å
D.9.96 Å
Solution
$\Delta E = 2.55\text{ eV = 13.6} \times 1^{2}\left( \frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}} \right)\text{ eV}$
$\therefore n_{1} = 2\text{ and }n_{2} = 4 $$$\text{Now, }\Delta\lambda\text{=3.32} \times \frac{4}{1} - 3.32 \times \frac{2}{1} = 6.64\overset{o}{A}$$
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