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The mass of an electron is `9.1 xx 10^(-31) kg`. If its K.E. is `3.0 xx 10^(-25)J`, calculate its wavelength |
Answer» Here, we are given: Kinetic energy, i.e., `(1)/(2) mv^(2) = 3.0 xx 10^(-25)J` `m = 9.1 xx 10^(-31) kg, h = 6.6 xx 10^(-34) kg m^(2) s^(-1)` `:. (1)/(2) xx (9.1 xx 10^(-31)) v^(2) = 3.0 xx 10^(25) or v^(2) = (3.0 xx 10^(-25) xx 2)/(9.1 xx 10^(-31)) = 659340 or v = 812 m s^(-1)` `:. lamda = (h)/(mv) = (6.626 xx 10^(-34) kgm^(2) s^(-1))/((9.1 xx 10^(-31) kg) xx 812 ms^(-1)) = 8967 xx 10^(-10) m = 896.7 nm` Alternatively, K.E. `= (1)/(2) mv^(2) or v = sqrt((2K.E.)/(m))` `lamda = (h)/(mv) = (h)/(m) xx sqrt((m)/(2K.E.)) = (h)/(sqrt(2m xx K.E.)) = (6.626 xx 10^(-34) kg m^(2) s^(-1))/(sqrt(2 xx 9.1 xx 10^(-31) kg xx 3.0 xx 10^(-25) j)) = 8967 xx 10^(-10)m` |
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