The wavelength of electron in the third orbit of hydrogen atom is \(6\pi a_0\). The kinetic energy of electron (in J) is equal to
\[
\left(
\text{Where, }
K=\frac{h^2}{\pi^2a_0^2m_e};
\;
a_0=\text{radius of first orbit of hydrogen};
\;
m_e=\text{mass of electron},
\;
h=\text{Planck's constant}
\right)
\]
Show Hint
For a particle,
\[
\boxed{
E_k=\frac{h^2}{2m\lambda^2}.
}
\]
Substitute the given de Broglie wavelength directly and simplify using the given constant.