Question #c8745
1 Answer
#lambda = "0.0123 nm"#
Assuming you mean
One electron-volt is the amount of energy needed to push an electron through a potential difference of one volt.
So, the energy involved in this process is
#K = 1/2 mv^2 = p^2/(2m)# (or
#p = sqrt(2mK)# )
#lambda = h/p = h/(mv)# where:
#h = 6.626 xx 10^(-34)# #"J"cdot"s"# is Planck's constant.#m# is the mass of the electron in#"kg"# .#v# is its speed in#"m/s"# .#p# is its momentum.#lambda# is the wavelength of the electron in#"m"# .
So, first, we convert the energy from
#10^4 cancel"eV" xx (1.602 xx 10^(-19) "J")/(cancel"1 eV") = 1.602 xx 10^(-15)# #"J"#
And now, if we derive
#lambda = h/(sqrt(2mK))#
#= (6.626 xx 10^(-34) "kg"cdot"m"^2"/s")/(sqrt(2cdot 9.109 xx 10^(-31) "kg" cdot 1.602 xx 10^(-15) "kg"cdot"m"^2"/s"^2)#
#= 1.227 xx 10^(-11)# #"m"#
A more useful unit here would be the
#color(blue)(lambda = "0.0123 nm")#
This is in the X-ray region of the EM spectrum.