Question

A proton is accelerated to 225V. Its de-Broglie wavelength is:

Answer 1

Answer:

The  value  is  $\lambda = 1.9109 *10^{-12} \ m$

Explanation:

From the question we are told that

  The  potential of the proton is  $V = 225 \ V$

Generally the momentum of the particle is mathematically represented as

         $p = \sqrt{ 2 * m * V * e }$

Here  e is the charge on the proton with value  

       $e = 1.60 *10^{-19} \ C$

      m is the mass of the proton with value  $m = 1.67 *10^{-27} \ kg$

So

    $p = \sqrt{ 2 * (1.67*10^{-27} ) * 225 * 1.6*10^{-19}}$

=>    $p = 3.4676 *10^{-22} \ kg \cdot m/ s$

So the de-Broglie wavelength isis mathematically represented as

     $\lambda = \frac{h}{p}$

Here  h is the Planck's  constant with value  

       $h = 6.626 *10^{-34} \ J\cdot s$

=>   $\lambda = \frac{6.626 *10^{-34}}{3.4676 *10^{-22} }$

=>$\lambda = 1.9109 *10^{-12} \ m$