Question

Calculate de-broglie wavelength of an electron whose kinetic energy is 50 ev.

Answer 1

de Broglie wavelength of a particle of mass, m

and linear momentum,p is given by

Lambda=h/p.

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Here,h is Planck's constant = 6.625x10"-34 J-s.

Now, Kinetic energy,K= p"2/(2m) OR

p=[2Km]"1/2

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2)

Putting this value in equation (1), we have

Lambda =h/[2Km]"1/2

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Given K=121eV=(121)(1.6)(10"-19) J

Mass of electron, m=(9.1)(10"-31)kg.

Therefore,with these values, equation (3)

becomes

Lambda =(6.625)(10"-34)/[(2)(121)(1.6)(10"-19)(9.1)

(10"-31)]

This yields Lambda =(1.116)(10"-10) m.

Answer 2

Given that,

Kinetic energy of the electrons, E = 50 eV

To find,

The De-Broglie wavelength of an electron.

Solution,

The De Broglie wavelength in terms of kinetic energy of an electron is given by the below relation as :

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

h is Planck's constant

m is mass of electron

$\lambda=\dfrac{6.63\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-19}\times 50\times 1.6\times 10^{-19}} }\\\\\lambda=1.73\times 10^{-16}\ m$

So, the De-Broglie wavelength of an electron is $1.73\times 10^{-16}\ m$.

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De-Broglie wavelength

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