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

User Billy Reilly
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1 Answer

15 votes
15 votes

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.

User Jaydo
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