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The kinetic energy of an electron is `5 eV`. Calculate the de - Broglie wavelength associated with it `( h = 6.6 xx 10^(-34) Js , m_(e) = 9.1 xx 10^(-31) kg)`A. `5.47 Å`B. `109 Å`C. `2.7 Å`D. None of these |
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Answer» Correct Answer - A `lambda = (h)/(sqrt(2 m E)) = (6.6 xx 10^(-34))/(sqrt(2 xx 9.1 xx 10^(-31) xx 5 xx 1.6 xx 10^(-19))` `= 5.469 xx 10^(-10) m = 5.47 Å` |
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