1.

Calculate the de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1 kV

Answer»

Solution :Energy acquired by the electron (as kinetic energy) after being accelerated by a POTENTIAL difference of 1 kV (i.e 1000 volts)
` = 100 eV`
` = 1000 xx 1.609 xx 10^(-19) , (1EV) = 1.609 xx 10^(-19) J)`
(Energy in joules = Charge on the electron in coulombs × Pot. diff. in volts)
` = 1.609 xx 10^(-16) J`
i.e. Kinetic energy
`(1/2 mv^2) = 1.609 xx 10^(-16) J`
`1/2 xx 9.1 xx 10^(-31) v^2 = 1.609 xx 10^(-16) J`
`v^2 = 3.536 xx 10^14`
` v = 1.88 xx 10^7 MS^(-1)`
` therefore lamda = (h)/(mv) = (6.626 xx 10^(-34) )/(9.1 xx 10^(-31) xx 1.88 xx 10^7)`
`= 3.87 xx 10^(-11) m`


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