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Calculate de Broglie wavelength of an electron travelling at `1 %` of the speed of light.A. `2.73 xx 10^-24`B. `2.4 xx 10^-10`C. `242.2 xx 10^10`D. None of these |
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Answer» Correct Answer - B (b) One percent of the speed of light is `v = ((1)/(100)) (3.00 xx 10^9 ms^-1)` =`(3.00 xx 10^8 ms^-1)` Momentum of the electron `(p) = mv` =`(9.11 xx 10^-31 kg)(3.00 xx 106 ms^-1)` =`2.73 xx 10^-24 kg ms^-1` The de-Broglie wavelength of this electron is `lamda = (h)/(p) = (6.626 xx 10^-34)/(2.73 xx 10^-24 kgms^-1)` `lamda = 2.424 xx 10^-10 m`. |
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