Question

The wavelength of light from the spectral emission line of sodium is $589nm$. Find the kinetic energy at which
(i) an electron and
(ii) a neutron would have the same de-Broglie wavelength.

• A. $K{E}_e=5\times 10^{-25}J,K{E}_n=6.5\times 10^{-28}J$
• B. $K{E}_e=6.96\times 10^{-25}J,K{E}_n=3.81\times 10^{-28}J$
• C. $K{E}_e=1.25\times 10^{-25}JK{E}_n=6.23\times 10^{-28}J$
• D. $K{E}_e=3.26\times 10^{+25}J,K{E}_n=4.06\times 10^{+28}J$

Answer 1

Answer: (B)

Given, wavelength of light $=589nm=589\times 10^{-9}m$
Mass of electron, $m_e=9.1\times 10^{-31}kg$
Mass of neutron, $m_n=1.67\times 10^{-27}kg$
Planck's constant, $h=6.62\times 10^{-34}J-s$
(i) Using of formula, $\lambda =\frac{h}{\sqrt{2KE{m}_e}}$
Kinetic energy of electron,
$K{E}_e=\frac{h^2}{2{\lambda }^2{m}_e}=\frac{{\left(6.63\times 10^{-34}\right)}^2}{2\times {\left(589\times 10^{-9}\right)}^2\times 9.1\times 10^{-31}}$
$=6.96\times 10^{-25}J$
(ii) Kinetic energy of neutron
$K{E}_n=\frac{h^2}{2{\lambda }^2{m}_n}=\frac{{\left(6.63\times 10^{-34}\right)}^2}{2\times {\left(589\times 10^{-9}\right)}^2\times 1.66\times 10^{-27}}$
$=3.81\times 10^{-28}J$