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(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.Mass of one $H_2$ molecule $(m)$ = $\frac{2 \, g/mol}{6.023 	imes 10^{

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$4$

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(B) The de-Broglie wavelength is given by the formula $\lambda = \frac{h}{mv}$.Mass of one $H_2$ molecule $(m)$ = $\frac{2 \, g/mol}{6.023 	imes 10^{23} \, mol^{-1}} \approx 3.32 	imes 10^{-24} \, g$.Given velocity $(v)$ = $5 	imes 10^4 \, cm \cdot sec^{-1}$.Planck's constant $(h)$ = $6.626 	imes 10^{-27} \, erg \cdot sec$.Substituting the values:$\lambda = \frac{6.626 	imes 10^{-27} \, erg \cdot sec}{(3.32 	imes 10^{-24} \, g) 	imes (5 	imes 10^4 \, cm \cdot sec^{-1})}$$\lambda = \frac{6.626 	imes 10^{-27}}{16.6 	imes 10^{-20}} \, cm \approx 3.99 	imes 10^{-8} \, cm$.Since $1 \, \mathring{A} = 10^{-8} \, cm$,the wavelength is approximately $4 \, \mathring{A}$.

If the velocity of a hydrogen molecule is $5 \times 10^4 \, cm \, sec^{-1}$,then its de-Broglie wavelength is ............. $\mathring{A}$.

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