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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. Given `h=6.6xx10^(-34)Js, m_(e)=9xx10^(-31)kg , 1eV=1.6xx10^(-19)J`.

A

725 pm

B

500 pm

C

322 pm

D

112 pm

Text Solution

Verified by Experts

The correct Answer is:
D

For a particle , de Broglie wavelength `lamda = h/p`
Kinetic energy , `K = p^2/(2m) :. Lamda =(h)/(sqrt(2mK))`
For and electron `lamda=h/(sqrt(2m_eK_e))`
`=((6.63 xx10^(-34)Js))/(sqrt(2xx(9.11 xx10^(-31)kg)xx(120 xx 1.6 xx10^(-19)J)))`
`= 112 xx 10^(-12) m = 112 "pm " [ :. 1 " pm" = 10^(-12)m]`
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