Question

If the shortest wavelengths of the continuous spectrum coming out of a Coolidge tube is $0.1\stackrel{\circ }{\text{A}}$, then the de Broglie wavelength of the electron reaching the target metal in the Coolidge tube is approximately $(hc=12400eV-\stackrel{\circ }{\text{A}},h=6.63\times {10}^{-34}\text{in MKS system and}\text{mass of electron}=9.1\times {10}^{-31}\text{kg})$

• A. $3.5\stackrel{\circ }{\text{A}}$
• B. $0.035\stackrel{\circ }{\text{A}}$
• C. $35\stackrel{\circ }{\text{A}}$
• D. $1\stackrel{\circ }{\text{A}}$

Answer 1

The correct option is B $0.035\stackrel{\circ }{\text{A}}$

Kinetic energy of the electron reaching the target metal = maximum kinetic energy of x-ray photon

$K=\frac{12400}{0.1}eV=124\times {10}^3\text{eV}K=124\times 1.6\times {10}^{-16}\text{J}$

de Broglie wavelength of the electron:
$\lambda =\frac{h}{\sqrt{2mK}}\lambda =\frac{6.63\times {10}^{-34}}{\sqrt{2\times 9.1\times {10}^{-31}\times 124\times 1.6\times {10}^{-16}}}\lambda =0.035\stackrel{\circ }{\text{A}}$