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What is the a. momentum, b. speed, and...

What is the
a. momentum, b. speed, and
c. de Broglie wavelength of an electron with kinetic energy of 120 eV.

Text Solution

Verified by Experts

Kinetic energy, `K.E = 120 eV = 120 xx 1.6 xx 10^(-19)`
`K = K.E = 1.92 xx 10^(-17)J`
(a) Momentum of an electron, `P = sqrt(2 mK)`
`P = sqrt(2 xx 9.1 xx 10^(-31) xx 1.92 xx 10^(-17))`
`P = 5.91 xx 10^(-24) "kg ms"^(-1)`
(b) Speed of an electron,
`v = (P)/(m) = (5.91 xx 10^(-24))/(9.1 xx 10^(-31)) = 6.5 xx 10^(6) ms^(-1)`
(c) de-Broglie wavelength,
`lambda = (h)/(P) = (6.6 xx 10^(-34))/(5.91 xx 10^(-24)) = 1.117 xx 10^(-10) = 0.112 xx 10^(-9)m`
`lambda = 0.112 m`
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