Question

The kinetic energy of an electron in a hydrogen atom is 3.4 eV. Calculate its de-Broglie wavelength.
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Answer 1

GIVEN :-

• Kinetic energy of an electron in a hydrogen atom = 3.4 eV

TO FIND :-

• de - Broglie wavelength

SOLUTION :-

 de - Broglie wavelength formula ,

          $\bf \lambda =\dfrac{h}{p}$

Where ,

 $\lambda$ = Wavelength

 h = Planck's constant

 P = momentum

We know that ,

  $\bf P = \sqrt{2} mE$

$\longrightarrow \sf \lambda = \dfrac{h}{\sqrt{2} mE}\\\\\longrightarrow \lambda = \dfrac{6.626\times 10^{-34}}{\sqrt{2}\times 9.1\times 10^{-31} \times 3.4 \times 1.6 \times 10^{-19}} \\\\\longrightarrow \lambda = 0.66 \times 10^{-9} m$

de - Broglie wavelength = $\sf 0.66 \times 10^{-9} \ m$

Additional information ,

In 1924, de - Broglie proposed that matter exhibit dual behaviour - particle and wave properties. He suggested that every object in motion possess a wave character but it is significant only for microscopic bodies.

de - Broglie's equation ,

   $\bf \lambda =\dfrac{h}{mv}$