Question

The wavelength of light from the spectral emission line of sodium is $$589 nm$$. Find the kinetic energy at which (a) electron and (b) neutron would have the same De-Broglie wavelength. Given that mass of neutron $$= 1.66 \times 10^{-27} kg$$.

Answer 1

Here, $$\lambda = 589 nm = 589 \times 10^{-9} m$$
The de Broglie wavelength of a particle having energy E is given by
$$\lambda = \dfrac{h}{\sqrt{2 m E}}$$
or $$E = \dfrac{h^2}{2 m\lambda^2}$$
a. For electron : Here, $$\lambda = 589 nm = 589 \times 10^{-9} m$$ and $$m = 9.1 \times 10^{-27} kg$$
$$\therefore E = \dfrac{h^2}{2 m \lambda^2} = \dfrac{(6.62 \times 10^{-34})^2}{2 \times 9.1 \times 10^{-31} \times (589 \times 10^{-9})^2}$$
$$= 6.94 \times 10^{-25} J$$
b. For neutron : Here, $$\lambda = 589 nm = 589 \times 10^{-9} m$$ and $$m = 1.66 \times 10^{-27} kg$$
$$\therefore E = \dfrac{h^2}{2 m \lambda^2} = \dfrac{(6.62 \times 10^{-34})^2}{2 \times 1.66 \times 10^{-27} \times (589 \times 10^{-9} )^2}$$
$$= 3.81 \times 10^{-28} J$$