The K.E. of an electron is 4.55 times 10^{-25 | vedclass

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Question

(A) Given: $K.E. = 4.55 	imes 10^{-25} \,J$,mass of electron $m = 9.1 	imes 10^{-31} \,kg$,Planck's constant $h = 6.626 	imes 10^{-34} \,J \cdot s$

Vedclass · Accepted Answer

$1.944 	imes 10^{-7} \,m$

Vedclass · Answer

(A) Given: $K.E. = 4.55 	imes 10^{-25} \,J$,mass of electron $m = 9.1 	imes 10^{-31} \,kg$,Planck's constant $h = 6.626 	imes 10^{-34} \,J \cdot s$. Using the relation $K.E. = \frac{1}{2}mv^2$,we find velocity $v = \sqrt{\frac{2 	imes K.E.}{m}} = \sqrt{\frac{2 	imes 4.55 	imes 10^{-25}}{9.1 	imes 10^{-31}}} = \sqrt{10^6} = 1000 \,m/s$. Using de Broglie's equation $\lambda = \frac{h}{mv}$,we get $\lambda = \frac{6.626 	imes 10^{-34}}{9.1 	imes 10^{-31} 	imes 1000} = \frac{6.626 	imes 10^{-34}}{9.1 	imes 10^{-28}} \approx 0.728 	imes 10^{-6} \,m = 7.28 	imes 10^{-7} \,m$. Wait,re-calculating: $\lambda = \frac{6.626 	imes 10^{-34}}{9.1 	imes 10^{-28}} = 0.7281 	imes 10^{-6} = 7.28 	imes 10^{-7} \,m$. Reviewing options,there seems to be a discrepancy. Let's re-check the calculation: $\lambda = \frac{6.626 	imes 10^{-34}}{9.1 	imes 10^{-31} 	imes 1000} = 7.28 	imes 10^{-7} \,m$. None of the options match exactly. Assuming a calculation error in the question source,the closest magnitude is $7.28 	imes 10^{-7} \,m$.

The $K.E.$ of an electron is $4.55 \times 10^{-25} \,J$. Calculate its $\lambda$.

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