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What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6 ×x 10^(–19) C) moving at a speed of 3 ×x10^7m/s in a magnetic field of 6 x× 10^(-4) T perpendicular to it?What is its frequency?Calculate its energy inkeV. (1eV = 1.6 xx 10^(-19) J) |
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Answer» Solution :Using eq. we FIND `r = mv// (qB) = 9 xx 10^(-31) kg xx 3 xx 10^7 ms^(-1) // (1.6 xx 10^(-19) C xx 6 xx 10^(-4) T)` ` = 28 xx 10^(-2) m = 28 cm` V `= v//(2pi r) = 17 xx 10^6 s^(-1) = 17 xx 10^6 Hz = 17 MHz` `E = (1//2) mv^2 = (1//2) 9 xx 10^(-31) kg xx 9 xx 10^14 m^2 // s62 = 40.5 xx 10^(-17) J` `= 4 xx 10^(-16) J = 2.5 KEV. ` |
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