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# An electron difraction experiment was performed with a beam of electrons accelerated by a potential difference of 10.0kV.What is the wavelength of the electron beam?(1KV = 1000V = 1000eV for the electron)

$(a)\;13.3pm\qquad(b)\;12.3pm\qquad(c)\;11pm\qquad(d)\;11.4pm$

$\lambda = \large\frac{h}{mV} = \large\frac{h}{\sqrt {2m(KE)}}$
KE = $10^4 eV = 10^4\times1.6\times10^{-19}J$
$\therefore \lambda = \large\frac{6.626\times10^{-34}}{\sqrt{2\times9.1\times10^{-31}\times1.6\times10^{-15}}}$
$=1.23\times10^{-11}m$
12.3 pm
Hence the anwer is (b)