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Electron starting from rest is acclerated between two point having p.d. of 20 V and 40 V. de-Broglie wavelength of electron…… |
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Answer» 0.75 Å `lambda=(H)/(sqrt(2meV))` `=(6.62xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx1.6xx10^(-19)xx20))` `0.274xx10^(-9)=2.75Å` |
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