Question

what is the wavelength, in pm, of an electron when its velocity is 66.67% the speed of light?

Answer #1

as we know the de brogile wavelength equation as lamda = plancks constant/momentum

λ = h/m*v

velocity of light is 3x
10^{8} m/s^{2} ; h = 6.636 * 10^{-34}
J.s

66.67% of speed of light= 66.67/100
* [3*10^{8}] = 2 *10^{8}

Mass of an electron =9.1
×10^{-31}kg

λ = 6.636*10^{-34}/ 9.1*10^{-31}*(2*10^{8})

we get lambda= 3.646 *
10^{-12} m

we also know that 1 m =
10^{12} pm

3.646 * 10^{-12} m = 3.646 *
10^{-12} x 10^{12} = 3.646 pm

**hence the final answer must
be 3.646 pm**

*Hope this helped you!*

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