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 $$$\text{Now, }\Delta\lambda\text{=3.32} \times \frac{4}{1} - 3.32 \times \frac{2}{1} = 6.64\overset{o}{A}$$
Create a free account to view solution
View Solution FreeMore Atomic Structure Questions
To what series does the spectral line of atomic hydrogen belong if its wave number is equal to the difference between t...The electrons identified by quantum numbers n and l,(i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l = 2 (iv) n = 3, l ...In the Rutherford scattering experiment, the number of alpha particles scattered at an angle θ = 60o is 36 per minute. T...The wave function of 1s orbital of H-atom is $\Psi = \frac{1}{\sqrt{\pi}}\left( \frac{1}{a_{0}} \right)^{3/2}e^{- r/a_{0...A molecule XY2 contains two σ, two π - bonds and one lone pair of electron in the valence shell of X. The arr...