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What is the de - Broglie wavelength associated with an electron, accelerated through a potential difference of 100 Volts ? |
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Answer» Solution :Accelerating potential `V = 100 V`. The de BROGLIE wavelength `lambda` is `lambda=h//p=(1.227)/(sqrt(V))=NM` `lambda=(1.227)/(sqrt(100))nm=0.123nm` The de Broglie wavelength ASSOCIATED with an electron in this case is of the order of X-ray WAVELENGTHS. |
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