Respuesta :
Explanation:
Given that,
The mean kinetic energy of the emitted electron, [tex]E=390\ keV=390\times 10^3\ eV[/tex]
(a) The relation between the kinetic energy and the De Broglie wavelength is given by :
[tex]\lambda=\dfrac{h}{\sqrt{2meE}}[/tex]
[tex]\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 1.6\times 10^{-19}\times 390\times 10^3}}[/tex]
[tex]\lambda=1.96\times 10^{-12}\ m[/tex]
(b) According to Bragg's law,
[tex]n\lambda=2d\ sin\theta[/tex]
n = 1
For nickel, [tex]d=0.092\times 10^{-9}\ m[/tex]
[tex]\theta=sin^{-1}(\dfrac{\lambda}{2d})[/tex]
[tex]\theta=sin^{-1}(\dfrac{1.96\times 10^{-12}}{2\times 0.092\times 10^{-9}})[/tex]
[tex]\theta=0.010^{\circ}[/tex]
As the angle made is very small, so such an electron is not useful in a Davisson-Germer type scattering experiment.
