Question

Relativistic corrections become necessary when the expression for the kinetic energy $\frac{1}{2}m{v}^2$, becomes comparable with $m{c}^2$, where m is the mass of the particle. At what de Broglie waxelength will relativistic corrections become important for an electron?

• A. $\lambda =lnm$
• B. $\lambda =l0nm$
• C. $\lambda =l{0}^{-1}nm$
• D. $\lambda =l{0}^{-4}nm$

Answer 1

Answer: (D)

$\lambda =l{0}^{-4}nm$

We know that the de Broglie's equation is written as:

$p=\frac{h}{\lambda }$

$mv=\frac{h}{\lambda }$

velocity of electrn, $v=\frac{h}{\left(m\lambda \right)}$

The value of plank's constant is $h=6.6\times {10}^{-34}Js$

and the mass of the electron is $m=9\times {10}^{-31}kg$

$\left(a\right)$

For ${\lambda }_1=10nm={10}^{-8}m$

$v_1=\frac{6.6\times {10}^{-34}}{(9\times {10}^{-31})\times {10}^{-8}}$

$=\frac{2.2}{3}\times {10}^5\approx {10}^{5}m/s$

$\left(b\right)$

For ${\lambda }_2={10}^{-1}nm={10}^{-10}m$

$v_2=\frac{6.6\times {10}^{-34}}{9\times {10}^{-31}\times {10}^{-10}}$

$\approx {10}^{7}m/s$

$\left(c\right)$

For ${\lambda }_3={10}^{-4}nm={10}^{-13}m$

$v_3=\frac{6.6\times {10}^{-34}}{9\times {10}^{-31}\times {10}^{-13}}$

$\approx {10}^{10}m/s$

$\left(d\right)$

For ${\lambda }_4={10}^{-6}nm={10}^{-15}m$

$v_4=\frac{6.6\times {10}^{-34}}{9\times {10}^{-31}\times {10}^{-15}}$

$\approx {10}^{12}m/s$

Since $v_3$ and $v_4$ are greater than velocity of light . So, the relativistic correction can occur in these speeds of the electron.

Thus, option (C) and (d) are required wavelengths.