Atomic StructureMedium
Question
De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to
(Given Bohr radius $= 0.052\text{ }nm$ )
Options
A.0.067 nm
B.0.67 nm
C.1.67 nm
D.2.67 nm
Solution
$r = 0.052n^{2}$
For $n = 2$
$${r = 0.052 \times 4 }{= 0.208\text{ }nm }{Mvr = \frac{nh}{2\pi} }{\lambda = \frac{h}{Mv} = \pi r }{= 3.14 \times 0.208\text{ }nm }{= 0.65317\text{ }nm }{\approx 0.67\text{ }nm}$$
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