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What is the wavelength for the electron accelerated by ` 1.0 xx 10^(4) `volts? |
Answer» Step I. Calculating of the velocity of electron Energy (kinetic energy ) of electron `=1.0 xx 10^(4)` volts. `=1.0 xx 10^(4) xx 1.6 xx 10^(-19)J=1.6 xx 10^(-15)J` `=1.6 xx 10^(-15)kg m^(2)s^(-2)` or `" " 1//2 m upsilon^(2)=1.6 xx 10^(-15)kg m^(2)s^(-2)` or `" " upsilon =((2xx1.6 xx10^(-15)kg m^(2)s^(-2))/(9.1 xx 10^(-31)kg))^(1//2)=5.93 xx 10^(7) ms^(-1)` Step II. calculation of the wavelength of electron According to de Broglie equation, `lambda =(h)/(m upsilon) , lambda = ((6.626 xx 10^(-34) kg m ^(2) s^(-1)))/((9.1 xx 10^(-31 ) kg) xx (5.93 xx 10^(7) ms^(-1)))=1.22 xx 10^(-11)m`. |
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