The wavelength of light from the spectral emission line of sodium is 589 nm.

Solution:

Wavelength of light of a sodium line, λ = 589 nm = 589 × 10−9 m

Mass of an electron, me= 9.1 × 10−31 kg

Mass of a neutron, mn= 1.66 × 10−27 kg

Planck’s constant, h = 6.6 × 10−34 Js

(a) For the kinetic energy K, of an electron accelerating with a velocity v, we have the relation:

$K=\frac{1}{2} m_{e} v^{2}$    ....(1)

We have the relation for de Broglie wavelength as:

$\lambda=\frac{h}{m_{e} v}$

$\therefore v^{2}=\frac{h^{2}}{\lambda^{2} m_{e}{ }^{2}}$     ...(2)

Substituting equation (2) in equation (1), we get the relation:

$K=\frac{1}{2} \frac{m_{e} h^{2}}{\lambda^{2} m_{e}{ }^{2}}=\frac{h^{2}}{2 \lambda^{2} m_{e}}$    ....(3)

$=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times 10^{-9}\right)^{2} \times 9.1 \times 10^{-31}}$

$\approx 6.9 \times 10^{-25} \mathrm{~J}$

$=\frac{6.9 \times 10^{-25}}{1.6 \times 10^{-19}}=4.31 \times 10^{-6} \mathrm{eV}=4.31 \mu \mathrm{eV}$

Hence, the kinetic energy of the electron is 6.9 × 10−25 J or 4.31 μeV.

 

(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:

$\frac{h^{2}}{2 \lambda^{2} m_{n}}$

$=\frac{\left(6.6 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times 10^{-9}\right)^{2} \times 1.66 \times 10^{-27}}$

$=3.78 \times 10^{-28} \mathrm{~J}$

$=\frac{3.78 \times 10^{-28}}{1.6 \times 10^{-19}}=2.36 \times 10^{-9} \mathrm{eV}=2.36 \mathrm{neV}$