Show that the de Broglie wavelength of a nonrelativistic electron accelerated fr

Question

Show that the de Broglie wavelength of a nonrelativistic electron accelerated from rest through a potential difference V is given by lamda = 1.228/sqrtVnm, where V is in volts. HELP! This is what we have on our sheet, but I don't fully understand what the teacher did. The answer is lamda = 1.228nm/sqrtV (which makes no sense to begin with).

the set up makes some sense.

1/2 mv^2 = eV solve for v

v = sqrt 2(eV)/m also p = mv = sqrt2meV (why?)

Then he moves on to this formula (which makes sense)

lamda = h/p = h/2meV = 6.63 x 10^-34 Js/2(9.11x10^-31kg)(1.6x10^-19C)sqrtv)

lamd = 1.228nm/sqrtv

Explanation / Answer

from energy conservation; 1/2 mv^2 = eV =>v = sqrt 2(eV)/m also p = mv = sqrt(2m^2*eV/m) =sqrt(2meV) wavelength =h/p = h/sqrt(2meV) => wavelength = 6.63 x 10^-34/sqrt[2*(9.11x10^-31kg)*(1.6x10^-19C)V] => wavelength = 1.228 *10^-9 m/sqrt(V)

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