If the velocity of the electron in Bohr’s first orbit is 2.19 × 106 ms–1,


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



According to de Broglie's equation,

$\lambda=\frac{h}{m v}$


$\lambda=$ wavelength associated with the electron

$h=$ Planck's constant

$m=$ mass of electron

$v=$ velocity of electron

Substituting the values in the expression of $\lambda$ :

$\lambda=\frac{6.626 \times 10^{-34} \mathrm{Js}}{\left(9.10939 \times 10^{-31} \mathrm{~kg}\right)\left(2.19 \times 10^{6} \mathrm{~ms}^{-1}\right)}$

$=3.32 \times 10^{-10} \mathrm{~m}=3.32 \times 10^{-10} \mathrm{~m} \times \frac{100}{100}$

$=332 \times 10^{-12} \mathrm{~m}$

$\lambda=332 \mathrm{pm}$

$\therefore$ Wavelength associated with the electron $=332 \mathrm{pm}$

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