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(C) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p = \sqrt{2mK}$ and $K$ is the kinetic energy.Thus,$\lambda = \frac

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$5.56 	imes 10^{-25} 	ext{ J}$

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(C) The de Broglie wavelength $\lambda$ is given by $\lambda = \frac{h}{p}$,where $p = \sqrt{2mK}$ and $K$ is the kinetic energy.Thus,$\lambda = \frac{h}{\sqrt{2mK}}$,which implies $K = \frac{h^2}{2m\lambda^2}$.Given: $h = 6.6 	imes 10^{-34} 	ext{ J s}$,$m = 9 	imes 10^{-31} 	ext{ kg}$,and $\lambda = 6600 	imes 10^{-10} 	ext{ m} = 6.6 	imes 10^{-7} 	ext{ m}$.Substituting the values:$K = \frac{(6.6 	imes 10^{-34})^2}{2 	imes 9 	imes 10^{-31} 	imes (6.6 	imes 10^{-7})^2}$$K = \frac{6.6^2 	imes 10^{-68}}{18 	imes 10^{-31} 	imes 6.6^2 	imes 10^{-14}}$$K = \frac{10^{-68}}{18 	imes 10^{-45}} = \frac{1}{18} 	imes 10^{-23} \approx 0.0556 	imes 10^{-23} 	ext{ J} = 5.56 	imes 10^{-25} 	ext{ J}$.Since the work done equals the kinetic energy gained,the work done is $5.56 	imes 10^{-25} 	ext{ J}$.

The work done to accelerate an electron from rest so that it can have a de Broglie wavelength of $6600 \text{ Å}$ is nearly (Planck's constant $= 6.6 \times 10^{-34} \text{ J s}$ and mass of electron $= 9 \times 10^{-31} \text{ kg}$)

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