Question

The kinetic energy of an electron is 5ev calculate the de broglie wavelength associated with it

Answer 1

Answer:

For an electron with KE = 1 eV and rest mass energy 0.511 MeV, the associated DeBroglie wavelength is 1.23 nm, about a thousand times smaller than a 1 eV photon. (This is why the limiting resolution of an electron microscope is much higher than that of an optical microscope.)

Answer 2

The de broglie wavelength associated with the electron is  5.47 A.

Explanation:

it is given that,

Kinetic energy of an electron, $E=5\ eV=5\times 1.6\times 10^{-19}\ J$

$E=8\times 10^{-19}\ J$

To find,

The De Broglie wavelength

Solution,

The relation between the kinetic energy and the wavelength of an electron is given by :

$\lambda=\dfrac{h}{\sqrt{2mK}}$

Where

h is the Planck's constant

m is the mass of electron

$\lambda=\dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 8\times 10^{-19}}}$

$\lambda=5.469\times 10^{-10}\ m$

or

$\lambda=5.47\ A$

So, the de broglie wavelength associated with the electron is  5.47 A.

Learn more,

De Broglie wavelength

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