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An electron and a photon each have a wavelength 1.00 nm. Find
(i) their momenta,
(ii) the energy of the photon and
(iii) the kinetic energy of electron.

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Given, `lambda = 1 mm = 10^(-9)m, h = 6.63xx10^(-34)J - s, c = 3xx10^(8)m//s`
`E_(e )=(P^(2))/(2m_(e ))=((6.63xx10^(-25))^(2))/(2xx9.1xx10^(-31)xx1.6xx10^(-19)) eV therefore E_(e )=1.51 eV`.

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