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Question

Find the de-Broglie wavelength of an electron with kinetic energy of $$120\ eV$$.


A
95 pm
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B
102 pm
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C
112 pm
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D
124 pm
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Solution

The correct option is C $$112\ pm$$
Given :   $$K = 120eV$$     

Mass of electron    $$m = 9.1\times 10^{-31}$$   kg 
   
de-Broglie wavelength      $$\lambda=  \dfrac{h}{\sqrt{2mK}}$$

$$\therefore$$      $$\lambda=  \dfrac{6.6\times 10^{-34}}{\sqrt{2\times 9.1\times 10^{-31}\times 120 \times 1.6\times 10^{-19}}}$$

Or    $$\lambda = \dfrac{6.6\times 10^{-34}}{59.1\times 10^{-25}} = 0.112\times 10^{-9}$$ $$m = 112$$  $$pm$$

Physics

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