Question

debroglie wavelength of electron accelerated through a potential difference of 1 kev​

Answer 1

Answer:

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Answer 2

Answer:

What is the de-Broglie wavelength of an electron that has been accelerated from rest through a potential difference of 1keV?

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The question isn’t quite right, potential difference is specified in V (volt) not eV (electron volt). Electron volt is a unit of energy not potential difference. So I’m assuming OP meant a potential difference of 1kV .

In that case, the energy of the electron is:

E=q×V

E=1.6×10−19×1000

E=0.00000000000000016J

Now, The kinetic energy of an electron with respect to its momentum is specified as:

E=p22m

Combining both the energy equations,

p22m=0.00000000000000016

p2=2.912×10−46

p=1.706×10−23kgm/s

Therefore, the de Broglie’s wavelength is given by,

λ=hp

So, the wavelength here is,

λ=6.62×10−341.706×10−23

λ=3.880×10−11m

λ=0.0388nm

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