Question

Find de-Broglie wavelength of the electron in first Bohr orbit.

Answer 1

We have

$mvr=\frac{nh}{2\pi }$

$\Rightarrow v=\frac{nh}{2\pi mr}=\frac{1\times 6.626\times {10}^{-26}}{2\times 3.142\times 9.1\times {10}^{-31}\times 0.529\times {10}^{-10}}$

$=2.18\times {10}^{6}m{s}^{-1}$

De-Broglie wavelength,

$\lambda =\frac{h}{mv}$

$=\frac{6.626\times 106-26}{9.1\times {10}^{-31}\times 2.18\times {10}^6}$

$=3.3\stackrel{o}{A}$.