Question

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

• A. $KE _{e}=5 \times 10^{-25} \,J , KE _{n}=6.5 \times 10^{-28}\, J$
• B. $KE _{e}=6.96 \times 10^{-25} \,J , KE _{n}=3.81 \times 10^{-28} \,J$
• C. $KE _{e}=1.25 \times 10^{-25} \,J KE _{n}=6.23 \times 10^{-28}\, J$
• D. $KE _{e}=3.26 \times 10^{+25}\,J , KE _{n}=4.06 \times 10^{+28}\, J$

Answer 1

Answer: (B)

Given, wavelength of light $=589 \,nm =589 \times 10^{-9} \, m$
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, $\lambda=\frac{h}{\sqrt{2 KEm _{e}}}$
Kinetic energy of electron,
$KE _{e} =\frac{h^{2}}{2 \lambda^{2} m_{e}}=\frac{\left(6.63 \times 10^{-34}\right)^{2}}{2 \times\left(589 \times 10^{-9}\right)^{2} \times 9.1 \times 10^{-31}}$
$=6.96 \times 10^{-25} \,J$
(ii) Kinetic energy of neutron
$KE _{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}} $
$=3.81 \times 10^{-28} \, J$