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2.21 The mass of an electron is 9.1\times 10^{-31}\ \textup{kg} If its K.E. is 3.0\times 10^{-25}\ \textup{J}, calculate its wavelength.

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The wavelength of an electron can be found by de Broglie's equation:

\lambda = \frac{h}{mv}

Given the K.E. of electron 3.0\times 10^{-25}\ \textup{J} which is equal to \frac{1}{2}mv^2.

Hence we get, 

\frac{1}{2}mv^2 = 3.0\times10^{-25}J

v = \sqrt{\frac{2KE}{m}} = \sqrt{\frac{2(3.0\times10^{-25}J)}{9.1\times10^{-31}kg}} = 811.579\ m/s

Hence the wavelength is given by,

\lambda = \frac{6.626\times10^{-34}Js}{(9.1\times10^{-31}kg)(811.579\ m/s)}

= 8.9625\times10^{-7}m

Posted by

Divya Prakash Singh

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896.7 nm

Posted by

Krishna shinde

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