Question

The kinetic energy of an electron revolving in an orbit of hydrogen atom is $\frac{E_0}{9}$ Where $E_0$ is the kinetic energy in first Bohr orbit. If the de-Broglie wavelength of the electron is $\lambda$ then the radius of the orbit will be

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

Answer 1

The correct option is A $\frac{3\lambda }{2\pi }$

$E=\frac{E_0}{n^2}=\frac{E_0}{9}\Rightarrow n=3$

So, the circumference

$C=2\pi r=n\lambda =3\lambda$
$\Rightarrow r=\frac{3\lambda }{2\pi }$