The de-Broglie wavelength associated with a h | vedclass

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(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.Here,$h = 6.63 	imes 10^{-34} \ J \cdot s$ (Planck's constant).The mas

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$0.66$

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(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.Here,$h = 6.63 	imes 10^{-34} \ J \cdot s$ (Planck's constant).The mass of a hydrogen molecule $(H_2)$ is $m = 2 	imes 1.67 	imes 10^{-27} \ kg = 3.34 	imes 10^{-27} \ kg$.The velocity is $v = 3 \ km/s = 3 	imes 10^3 \ m/s$.Substituting these values into the formula:$\lambda = \frac{6.63 	imes 10^{-34}}{3.34 	imes 10^{-27} 	imes 3 	imes 10^3} \ m$.$\lambda = \frac{6.63 	imes 10^{-34}}{10.02 	imes 10^{-24}} \ m \approx 0.66 	imes 10^{-10} \ m$.Since $1 \ \mathring{A} = 10^{-10} \ m$,we get $\lambda = 0.66 \ \mathring{A}$.

The de-Broglie wavelength associated with a hydrogen molecule moving with a thermal velocity of $3 \ km/s$ will be ............ $\mathring{A}$.

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