Question

In H-atom, if ‘x’ is the radius of the first Bohr orbit, de Broglie wavelength of an electron in 3rd orbit is :

• A. 3 $\pi x$
• B. $6\pi x$
• C. $\frac{9x}{2}$
• D. $\frac{x}{2}$

Answer 1

Answer: (B)

$6\pi x$

Answer 2

$r_{1}\text{ = 0.529 }\text{Å}$

$r_{3}\text{ = 0.529 }\text{×}\text{ }(3)^{2}\text{ }\text{Å}\text{ = 9x}$

so, $\lambda = \frac{2\pi r}{n} = \frac{2\pi(9x)}{3} = 6\pi x.$