Question #3da69

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Answer 1 · 2017-02-21T16:31:04

#sf(lambda=1.23color(white)(x)nm)#

1 electron volt = #sf(1.6xx10^(-19)color(white)(x)J)#

We can equate this to the kinetic energy of the electron.

#:.##sf(1/2mv^2=1.6xx10^(-19)color(white)(x)J)#

The mass of the electron (ignoring relativistic effects) =#sf(9.1xx10^(-31)color(white)(x)"kg")#

#:.##sf(v^2=(2xx1.6xx10^(-19))/(9.1xx10^(-31))=0.35xx10^(12))#

#sf(v=sqrt(0.35xx10^(12))=5.91xx10^(5)color(white)(x)"m/s")#

Now we use the de Broglie expression:

#sf(lambda=h/(mv)#

#:.##sf(lambda=(6.63xx10^(-34))/(9.1xx10^(-31)xx5.91xx10^5)color(white)(x)m)#

#sf(lambda=1.23xx10^(-9)color(white)(x)m)#

#sf(lambda=1.23color(white)(x)nm)#