Question

write the de broglie wavelength of kinetic energy E

Answer 1

What is the De Broglie wavelength of an electron given it has a kinetic energy of 1 eV? You are given the mass of an electron is 9.11x10^-31 kg and Planck's constant is 6.63x10^-34


The De Broglie wavelength equation is as follows:

λ=h/p

We know the value of Planck's constant h and so to calculate the wavelength all we need is the momentum, which is equal to mv.

The kinetic energy is given as 1 eV. Remember 1 eV is equal to 1.6 x 10-19 Joules. Using the equation for kinetic energy and the given mass of the electron we can determine the velocity of the electron as follows:

K.E = 0.5mv2

Which can be rearranged to be in terms of velocity v:

v = (2K.Em)0.5

By substituting in 1.6 x 10-19 for K.E and 9.11 x 10-31 for m we get v = 5.93 x 105 ms-1 (remember to keep the full non-rounded value in your calculator!)

Then using the initial equation for the wavelength and remembering p = mv, we can substitute in our values for h, m and v as follows:

λ = 6.33 x 10-34 / (9.11 x 10-31 x 5.93 x 105)  

λ = 1.17 x 10-9 m

Answer 2

Explanation:

The formula of the D-Broglie wavelength is given by :

$\lambda=\dfrac{h}{mv}$

Here,

h is Planck's constant

m is mass of electron

v is speed of electron

We know that m times v is called momentum of an object. We know that the relation between kinetic energy and momentum is given by :

$E=\dfrac{p^2}{2m}$

E is kinetic energy

so,

$p=\sqrt{2mE}$

De-Broglie wavelength becomes,

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

Hence, this is the required solution.

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