Question

an electron is moving with a kinetic energy of 2.275 x 10⁻²⁵ joule. Calculate its de Broglie wavelength. (Mass of e⁻=9.1×10⁻³¹kg & h= 6.6×10⁻³⁴j/s)

Answer 1

We know that lambda (wavelength)=h/p
=h/m x v ,But we have kinetic energy = 2.2 ×10^ -25
An K E = 1/2 mv^2
Therefore putting values v= 7 x 10^2
And therefore putting values in h/m x v
We get lambda = 1 x 10^-6

Answer 2

Answer: The de-Broglie wavelength of an electron is $1.025\mu m$

Explanation:

De-Broglie wavelength is calculated by using the formula:

$\lambda=\frac{h}{mv}$      .....(1)

where,

$\lambda$ = wavelength of electron

h = Planck's constant = $6.6\times 10^{-34}Js$

m = mass of electron = $9.1\times 10^{-31}kg$

v = velocity of electron (to be calculated from kinetic energy)

Formula used to calculate kinetic energy of an electron is given as:

$E=\frac{1}{2}mv^2$

Putting values of energy and mass in above formula, we get:

$2.275\times 10^{-25}=\frac{1}{2}\times 9.1\times 10^{-31}\times v^2\\\\v=707.1m/s$

Now, putting all the values in equation 1, we get:

$\lambda=\frac{6.6\times 10^{-34}kgm^2s^2/s}{9.1\times 10^{-31}kg\times 707.1m/s}\\\\\lambda = 1.025\times 10^{-6}m=1.025\mu m$

Hence, the de-Broglie wavelength of an electron is $1.025\mu m$