Answer to Question #114899 in Quantum Mechanics for chandan

Question

Find the deBroglie wave length of (a)an electron whose speed is1.0×108m/s,and (b)an electron whose speed is2.0×108m/s.

Answer 1

By definition, the deBroglie wavelength is given by:

$\lambda = \dfrac{h}{p}$ , where $h = 6.6\cdot 10^{-34} J\cdot s$ is the Planck constant and $p$ is the momentum of the electron. In the relativistic case:

$p = \dfrac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$ , where $m = 9.1\cdot 10^{-31} kg$ is the electon mass, $c = 3\cdot 10^8$ speed of light in vacuum and $v$ is the electron speed.

Finaly obtain:

$\lambda = \dfrac{h}{mv}\sqrt{1-\frac{v^2}{c^2}}$ .

Substitute numerical values:

a) $\lambda = \dfrac{6.6\cdot 10^{-34}}{9.1\cdot 10^{-31}\cdot 1\cdot10^8}\sqrt{1-\frac{1\cdot10^{16}}{9\cdot 10^{16}}} = 6.84\cdot 10^{-12} m$

b) $\lambda = \dfrac{6.6\cdot 10^{-34}}{9.1\cdot 10^{-31}\cdot 2\cdot10^8}\sqrt{1-\frac{4\cdot10^{16}}{9\cdot 10^{16}}} = 2.7\cdot 10^{-12} m$