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Find the de-Broglie wavelength of an electron with kinetic energy of 120 eV. |
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Answer» SOLUTION :`lambda=(H)/(sqrt(2mE))` `=(6.625 XX 10^(-34))/(sqrt(2 xx 9.1 xx 10^(31) xx 120 xx 1.6 xx 10^(-19)))` `=1.121 xx 10^(-10)`m. |
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