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The mass of an electron is `9.1 xx 10^(-31)J`, if its K.E. is `3.0 xx 10^(-25)J`. Calculate its wavelength. |
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Answer» Since `K.E = (1)/(2)mv^(2)` `v = ((2K.E)/(m)) = ((2xx 3.0xx 10^(-25)kg m^(2)s^(-2))/(9.1 xx 10^(-31)))^(1/2) =812 ms^(-1)` `lambda = (h)/(mv) = (6.626 xx 10^(-34)Js)/((9.1 xx 10^(-31)kg)(813 ms^(-1)))` `= 8967 xx 10^(-10)m = 896.7 nm` |
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