4)if the potential difference applied to a coolidge tube is 28.8kV,find out the maximum kinetic energy energy and velocity of the electrons hitting the target.
ans 4.61×10^-15,1.01×10^8 m/s

KE = 12mv2Ke eV = 1.6×10-19×28.8×1000 = 4.6×10-15 Jv = 2×KEm = 2×28.8×1000×1.69.1×10-31 = 1.01×108 m/s

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