Calculate de broglie wavelength of revolving electron of li^2+
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De broglie wavelength
lambda =h/mv
Li^2+ ion,
lamba=hn/m X 2.187x10^6 m/s X Z
putting, h =6.626 x10^-34 kgm^2/s
m=9.1 x 10^-31 kg
Z=3(Li)
n=1(not mentioned)
solving this, we get
wavelength=1.257 x 10^-10 m
=1.257 Angstrom
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