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The equation \lambda=\frac{1.227}{x} \mathrm{~nm} can be used to find the de-Brogli wavelength of an electron. In this equation \mathrm{x} stands for :
Where \mathrm{m}= mass of electron
          \mathrm{P}= momentum of electron
         \mathrm{K}=Kinetic energy of electron
         \mathrm{V}=Accelerating potential in volts for electron
 

Option: 1

\sqrt{\mathrm{mK}}


Option: 2

\sqrt{\mathrm{P}}


Option: 3

\sqrt{\mathrm{K}}


Option: 4

\sqrt{\mathrm{V}


Answers (1)

best_answer

\mathrm{\lambda=\frac{1.227}{x} \mathrm{~nm}}
For electron, de-broglie wavelength is

\mathrm{\lambda =\frac{12.27}{\sqrt{V}} \AA }

    \mathrm{=\frac{1.227}{\sqrt{V}} \mathrm{~nm} }

\mathrm{x =\sqrt{V}}

Hence 4 is correct option




 

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