If the velocity of the electron in Bohr’s first orbit is $$2.19 \times 10^{6} ms^{–1}$$ , calculate the de Broglie wavelength associated with it.

Asked by Abhisek | 1 year ago |  133

##### Solution :-

As per de Broglie’s equation,

$$\lambda =\dfrac{h}{m\nu }$$

Where, λ is the wavelength of the electron

h is Planck’s constant

m is the mass of the electron

v denotes the velocity of electron

Substituting these values in the expression for λ:

$$\lambda =\dfrac{h}{m\nu }\\ \\ =\dfrac{(6.626\times 10^{-34})}{9.103939\times 10^{-31}kg(2.19\times 10^{6}ms^{-1})}\\ \\$$

$$=3.32\times 10^{-10}m\lambda =332pm$$

Answered by Abhisek | 1 year ago

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