If the velocity of the electron in Bohr's fir | vedclass

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} 	ext{ J s}$$m = 9.10938 	i

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$332$

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(A) The de Broglie wavelength $\lambda$ is given by the formula $\lambda = \frac{h}{mv}$.Given:$h = 6.626 	imes 10^{-34} 	ext{ J s}$$m = 9.10938 	imes 10^{-31} 	ext{ kg}$$v = 2.19 	imes 10^6 	ext{ m s}^{-1}$Substituting the values:$\lambda = \frac{6.626 	imes 10^{-34}}{(9.10938 	imes 10^{-31}) 	imes (2.19 	imes 10^6)}$$\lambda = \frac{6.626 	imes 10^{-34}}{19.9495 	imes 10^{-25}}$$\lambda \approx 0.33214 	imes 10^{-9} 	ext{ m}$$\lambda \approx 332.14 	imes 10^{-12} 	ext{ m}$Since $1 	ext{ pm} = 10^{-12} 	ext{ m}$, the wavelength is approximately $332 	ext{ pm}$.

If the velocity of the electron in Bohr's first orbit is $2.19 \times 10^6 \text{ m s}^{-1}$, calculate the de Broglie wavelength associated with it. [$h = 6.626 \times 10^{-34} \text{ J s}$ and mass of electron = $9.10938 \times 10^{-31} \text{ kg}$] (in $\text{ pm}$)

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