Wavelength of light of a sodium line,
\(\begin{array}{l}\lambda\end{array} \)
-9Mass of an electron, me = 9.1 x 10-31 Kg
Mass of a neutron, mn = 1.66 x 10-27 Kg
Planck’s constant, h = 6.6 x 10-34 Js
(a) For the kinetic energy, K, of an electron accelerating with the velocity v, we have the relation:
K =
\(\begin{array}{l}\frac{1}{2}mv^{2}\end{array} \)
(1)We have the relation for de Broglie wavelength as:
\(\begin{array}{l}\lambda = \frac{h}{m_{e}v}\end{array} \)
Therefore,
\(\begin{array}{l}v^{2} = \frac{h^{2}}{\lambda ^{2}m_{e}^{2}}\end{array} \)
(2)Substituting equation (2) in equation (1), we get the relation:
K =
\(\begin{array}{l}\frac{1}{2}\end{array} \)
\(\begin{array}{l}\frac{m_{e}h^{2}}{\lambda ^{2}m_{e}^{2}}\end{array} \)
=
\(\begin{array}{l}\frac{h^{2}}{2\lambda^{2}m_{e} }\end{array} \)
(3)=
\(\begin{array}{l}\frac{(6.6\times 10^{-34})^{2}}{2\times (589\times 10^{-9})^{2}\times 9.1\times 10^{-31}}\end{array} \)
= 6.9 x 10-25 J =
\(\begin{array}{l}\frac{6.9\times 10^{-25}}{1.6\times 10^{-19}}\end{array} \)
= 4.31 x 10-6 eV Hence, the kinetic energy of the electron is 6.9 x 10-25 J.
(b) Using equation (3), we can write the relation for the kinetic energy of the neutron as:
=
\(\begin{array}{l}\frac{h^{2}}{2\lambda ^{2}m_{n}}\end{array} \)
=
\(\begin{array}{l}\frac{(6.6\times 10^{-34})^{2}}{2\times (589\times 10^{-9})^{2}\times 1.66\times 10^{-27}}\end{array} \)
= 3.78x 10 -28 J
=
\(\begin{array}{l}\frac{3.78\times 10^{-28}}{1.6\times 10^{-19}}\end{array} \)
= 2.36 x 10-9 eV = 2.36 neV
Therefore, the neutron has a kinetic energy of 3.78 x 10-28 J or 2.36 neV.