Calculate the wavelength of an electron moving with a velocity of 2.05 × 107 m s–1

Asked by Abhisek | 1 year ago |  92

#### 1 Answer

##### Solution :-

As per de Broglie’s equation,

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

Where, λ denotes thr wavelength of the moving particle

m is the mass of the particle

v denotes the velocity of the particle

h is Planck’s constant

Substituting these values in the expression for λ:

$$\lambda =\dfrac{(6.626\times 10^{-34})Js}{(9.10939\times 10^{-31}kg)(2.05\times 10^{7}ms^{-1})}\\ \\ =3.548\times 10^{-11}\, m$$

Therefore, the wavelength associated with the electron which is moving with a velocity of

$$2.05\times 10^{7}\, ms^{-1}\: is\:\: 3.548\times 10^{-11}\, m$$

Answered by Abhisek | 1 year ago

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