an electron beam from an acclerator is with kinetic energy 1.6*10-17 j what is its debroglie wavelength
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Answer:
de broglie wavelength=h/mv
Explanation:
=6.67×10^-34/p[p=mv]
=6.67 ×10^-34/√2mK.E [square root of 2m K.E]
=6.67×10^-34/√2×9.1×10^-31×1.6×10^-17[square root of 2×9.1×10^-31×1.6×10^-17]
=6.67 ×10^-33/5.40 ×10-24
=1.23 ×10^-10
=1.23A°
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