3 years ago

I need help asappp!!! Help b trcrcrhhvvcc

Physics

1 answer:

ozzi3 years ago

3 0

Answer:

the answer is c. hope this help you out

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Answer : The de-Broglie wavelength of this electron, $0.101\AA$

Explanation :

The formula used for kinetic energy is,

$K.E=\frac{1}{2}mv^2$ ..........(1)

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$\lambda=\frac{h}{mv}$

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Now put all the given values in the above formula (3), we get:

$\lambda=\frac{6.626\times 10^{-34}Js}{\sqrt{2\times 9.11\times 10^{-31}Kg\times 4.71\times 10^{-15}J}}$

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Therefore, the de-Broglie wavelength of this electron, $0.101\AA$

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