Question

If radius of $3^{rd}$ orbit of $L{i}^{2+}$ ion is 'x' cm, then de-broglie wavelength of electrons in the $1^{st}$ orbit is :

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

Answer 1

Answer: (A)

$\frac{2\pi x}{3}$ cm

Here $n=3;z=3$
$r_3=r_1\times \frac{n^2}{z}$
So,$r_1=\frac{r_3}{3}=\frac{x}{3}$cm
De-broglie's wavelength$=2\pi r_1=\frac{2\pi x}{3}$ cm