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(A) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Given:Mass $m = 9.109 	imes 10^{-31} \ kg$ (No

Vedclass · Accepted Answer

$6.044 	imes 10^{-9}$

Vedclass · Answer

(A) According to the de Broglie equation,the wavelength $\lambda$ is given by $\lambda = \frac{h}{mv}$.Given:Mass $m = 9.109 	imes 10^{-31} \ kg$ (Note: The original question had $10^{-28} \ g$,which is $10^{-31} \ kg$).Velocity $v = 1.20 	imes 10^5 \ m/s$.Planck's constant $h = 6.626 	imes 10^{-34} \ J \cdot s$.Substituting the values:$\lambda = \frac{6.626 	imes 10^{-34}}{(9.109 	imes 10^{-31}) 	imes (1.20 	imes 10^5)}$$\lambda = \frac{6.626 	imes 10^{-34}}{10.9308 	imes 10^{-26}}$$\lambda \approx 0.6062 	imes 10^{-8} \ m = 6.062 	imes 10^{-9} \ m$.Comparing with the given options,the closest value is $6.044 	imes 10^{-9}$.

The wavelength associated with a particle of mass $9.109 \times 10^{-31} \ kg$ moving with a velocity of $1.20 \times 10^5 \ m/s$ is ....... meters.

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