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An electron is revolving around the nucleus with a constant speed of 2.2xx10^(6)ms^(-1). Find the de-Broglie wavelength associated with it. |
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Answer» Solution :Here `v=2.2xx10^(6)ms^(-1)`, mass of electron `m=9.1xx10^(-31)kg and h=6.63xx10^(-34)JS` `therefore`de-Broglie wavelength `lamda=(h)/(MV)=(6.63xx10^(-34))/(9.1xx10^(-31)xx2.2xx10^(6))=3.31xx10^(-10)m=3.31` Å |
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