Question

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

• A. $3\pi x$
• B. $6x$
• C. $6\pi x$
• D. $\frac{x}{2}$

Answer 1

Answer: (C)

Angular momentum is quantized, hence

$Mv{r}_n=\frac{nh}{2\pi }$

By Bohr's theory $(a_0=$ Bohr's radius $=x)$

$r_n=\frac{n^2{a}_0}{Z}=\frac{9a_0}{1}=9x$

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

$\therefore mv=\frac{h}{6\pi x}$

By de-Broglie's equation

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