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An electron is accelerated through a potential difference of 81V. What is the de Broglie wavelength associated with it? To which part of electromagnetic spectrum does this wavelength corresspond ? |
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Answer» Solution :de-Broglie wavelength of an ELECTRON beam accelerated through a POTENTIAL difference of V volts is `lambda = (h)/(SQRT(2meV)) = (1.23)/(sqrt(V)) nm` V = 81 V, so `lambda = (1.23)/(sqrt(81)) xx 10^(-9)m = 0.1366 xx 10^(-9)m` `lambda = 1.36 Å` X-ray is the part of ELECTROMAGNETIC spectrum does this wavelength corresponds. X-ray has the wavelengths ranging from about `10^(-8)` to `10^(-12)` m. |
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