Question

Calculate the De brogile wavelenth a) an electron with a kinetic energy of 100 eV, (b) a proton with a kinetic energy of 100 eV, (c) an electron in the first Bohr orbit of a hydrogen atom.

Answer 1

debroglie wavelength lambda= h/(2Em)^1/2

h(planck's constant)=6.626

Answer 2

Debroglie wavelength is

$\lambda=\frac{h}{p}=\frac{h}{\sqrt{2mE}}$

(a) for electron with KE-100eV

$\lambda=\frac{6.6*10^{-34} }{\sqrt{2*100*1.6*10^{-19}*9.1*10^{-31}}}$

$\lambda=1.22*10^{-10}m$

(b) for proton with KE-100eV

$\lambda=\frac{6.6*10^{-34} }{\sqrt{2*100*1.6*10^{-19}*1.67*10^{-27}}}$

$\lambda=0.2855*10^{-11}m$

(c)

for first orbit

$\lambda=2\pi r=2\pi\frac{ n ^{2} a_{0}}{Z}$

for hydrogen atom Z=1 and for first orbit n=1 a₀=0.0529nm

$\lambda=2\pi a_{0}$

$\lambda=2*3.14*0.0529*10^{-9}$

$\lambda=3.32*10^{-10}m$