If the kinetic energy of an electron of mass | vedclass

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(C) The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2 	imes K.E.}{m}}$.Substituting the values: $v = \sqrt{\frac

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$552.2$

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(C) The kinetic energy $(K.E.)$ is given by $K.E. = \frac{1}{2}mv^2$,so $v = \sqrt{\frac{2 	imes K.E.}{m}}$.Substituting the values: $v = \sqrt{\frac{2 	imes 8.0 	imes 10^{-25}}{9.0 	imes 10^{-31}}} = \sqrt{\frac{16}{9} 	imes 10^6} = \frac{4}{3} 	imes 10^3 \ m/s$.The de Broglie wavelength is given by $\lambda = \frac{h}{mv}$.Substituting $h = 6.626 	imes 10^{-34} \ J \cdot s$,$m = 9.0 	imes 10^{-31} \ kg$,and $v = \frac{4}{3} 	imes 10^3 \ m/s$:$\lambda = \frac{6.626 	imes 10^{-34}}{9.0 	imes 10^{-31} 	imes (4/3) 	imes 10^3} = \frac{6.626 	imes 10^{-34}}{12 	imes 10^{-28}} = 0.55216 	imes 10^{-6} \ m$.Converting to $nm$: $\lambda = 0.55216 	imes 10^{-6} 	imes 10^9 \ nm = 552.16 \ nm \approx 552.2 \ nm$.

If the kinetic energy of an electron of mass $9.0 \times 10^{-31} \ kg$ is $8.0 \times 10^{-25} \ J$,the wavelength of this electron in $nm$ is

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