Question

Calculate the wavelength of an electron moving with a velocity of $2.05 \times {10^7}m{s^{ - 1}}$.

Answer 1

The wavelength of the electron can be calculated using de Broglie’s equation $\lambda = \dfrac{h}{{m\nu }}$. This equation is applicable to any material particle.
Formula used:
$\lambda = \dfrac{h}{{m\nu }}$

Complete step by step answer:
The significance of de Broglie’s equation lies in the fact that it relates the particle characters (e.g. mass) with the wave character (e.g. wavelength) of matter.
According to the de Broglie’s equation,
$\lambda = \dfrac{h}{{m\nu }}$
where, $m =$mass of the particle in $kg$
$\nu =$velocity of the particle in $m{s^{ - 1}}$
$h =$Planck’s constant $= 6.626 \times {10^{ - 34}}joule\,\sec \,$or $kg{m^2}{s^{ - 1}}$
$\lambda =$de Broglie’s wavelength in metres
It is given that $\nu =$ $2.05 \times {10^7}m{s^{ - 1}}$and we know that the mass of an electron $(m) =$ $9.1 \times {10^{ - 31}}kg$
Now, substituting the given values in the de Broglie’s equation, we get,
$\lambda = \dfrac{h}{{m\nu }} \\\ \Rightarrow \lambda = \dfrac{{6.626 \times {{10}^{ - 34}}kg{m^2}{s^{ - 1}}}}{{9.1 \times {{10}^{ - 31}}kg\,\,\, \times \,\,2.05 \times {{10}^7}m{s^{ - 1}}\,}} \\\ \Rightarrow \lambda = 3.552 \times {10^{ - 11}}m \\\$
Hence, the wavelength of an electron moving with a velocity of $2.05 \times {10^7}m{s^{ - 1}}$is $3.552 \times {10^{ - 11}}m$.

Additional information:
The S.I unit of Planck’s constant $(h)$: is $joule\,\sec \,$or $kg{m^2}{s^{ - 1}}$, where, $1\,Joule = 1kg{m^2}{s^{ - 2}}$.
$E = h\nu \\\ \Rightarrow h = \dfrac{E}{\nu } = \dfrac{J}{{{{\sec }^{ - 1}}}} = J\,\,\sec \\\$
$Now,\,\,\lambda = \dfrac{h}{{m\nu }}\,\, \\\ or,\,h = \lambda m\nu = (m)(kg)(m{s^{ - 1}}) = kg{m^2}{s^{ - 1}} \\\ \\\ As,\,\,E = h\nu \, \\\ \Rightarrow \,Energy = (kg{m^2}{s^{ - 1}})({s^{ - 1}}) = kg{m^2}{s^{ - 2}} \\\ so,\,1\,J = 1\,kg{m^2}{s^{ - 2}} \\\$

Note:
The de Broglie’s equation is applicable to any material particle, but it has significance only in the case of microscopic particles.