Question

The wavelength of light from the spectral emission line of sodium is 589 nm. Find the kinetic energy at which

(a) an electron, and

(b) a neutron would have the same de Broglie wavelength.

Answer 1

(a) $\lambda =589\times {10}^{-9}m$

$K=m{v}^2/2-\left(1\right);m$ is the mass of the electron.

We have de broglie wavelength as,

$\lambda =h/mv\Rightarrow v=h/m\lambda -\left(2\right)$

From $\left(1\right)and\left(2\right)$, we get $K=\frac{h^2}{2{\lambda }^2{m}_e}=6.9\times {10}^{-25}J=4.31\mu eV$

(b) $\lambda =589\times {10}^{-9}$

$K=m{v}^2/2-\left(1\right);m$ is the mass of neutron.

We have de broglie wavelength as,

$\lambda =h/mv\Rightarrow v=h/m\lambda -\left(2\right)$

From $\left(1\right)and\left(2\right)$, we get $K=\frac{h^2}{2{\lambda }^2{m}_e}=2.36\times {10}^{-9}eV=2.36neV$