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Q. If the radius of the first Bohr orbit is $x,$ then de Broglie wavelength of the electron in the third orbit is nearly

Structure of Atom

Solution:

Angular momentum is quantized, hence

$Mvr _{ n }=\frac{n h}{2 \pi}$

By Bohr's theory $\left(a_{0}=\right.$ Bohr's radius $\left.=x\right)$

$r_{n}=\frac{n^{2} a_{0}}{Z}=\frac{9 a_{0}}{1}=9 x $

$\therefore m v(9 x)=\frac{3 h}{2 \pi} $

$\therefore m v=\frac{h}{6 \pi x}$

By de-Broglie's equation

$\lambda=\frac{h}{m v}=\frac{h}{h / 6 \pi x}=6 \pi x$