Question

De-Broglie wavelength of electron in $nth$ orbit of $H-$atom is $\left(3.18\pi \right)A$. The value of $n$ is:

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is A $1$

Since,

$2\pi r=n\lambda$

r is the radius of Bohr orbit, $\lambda$ de Broglie wave length, n number of orbit.

Here,

$\lambda =3.18\stackrel{o}{A}$

$r=0.529\stackrel{o}{A}$

Therefore

$n=\frac{2\pi r}{\lambda }$

$=\frac{2\times 3.14\times 0.529}{3.18}=1.04$$\simeq 1$