The de Broglie wavelength (in nanometer) for an electronmoving with 18.72 x 104 m/s is:

Billy Reilly asked Dec 28, 2023

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Billy Reilly asked Dec 28, 2023

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The Broglie wave equation is

$\lambda=(h)/(p)$

Where p is the momentum (m*v) and h = 6.63x10^-34 J*s.

Find the momentum of an electron first.

$p=9.11*10^(-31)\operatorname{kg}\cdot18.72*10^4\cdot(m)/(s)=1.71*10^(-25)\operatorname{kg}\cdot(m)/(s)$

Then, find the Broglie wavelength.

$\lambda=\frac{6.63*10^(-34)J\cdot s}{1.71*10^(-25)\operatorname{kg}\cdot(m)/(s)}=3.88*10^(-9)m$

Therefore, the Broglie wavelength is 3.88x10^-9 meters.

Jaydo answered Jan 1

by Jaydo

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