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An electron of a hydrogen like atom is in excited, state. If total energy of the electron is -4.6 eV, then evaluate (i) the kinetic energy and (ii) the de-Brogli wavelength of the electron . |
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Answer» Given ,total energy of the electron `E=-4.6 EV` (i) Kinetic energy of electron K=-(Total energy E,) `rArr " " =-E=-(-4.6)=4.6 eV` (ii) de- Brogile wavelength `lambda_(d)=(h)/(sqrt(2 mK))=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx4.6xx1.6xx10^(-19)))` `0.57xx10^(-9)m=0.57 nm` |
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