Answer:
The value is [tex]\lambda = 1.9109 *10^{-12} \ m[/tex]
Explanation:
From the question we are told that
The potential of the proton is [tex]V = 225 \ V[/tex]
Generally the momentum of the particle is mathematically represented as
[tex]p = \sqrt{ 2 * m * V * e }[/tex]
Here e is the charge on the proton with value
[tex]e = 1.60 *10^{-19} \ C[/tex]
m is the mass of the proton with value [tex]m = 1.67 *10^{-27} \ kg[/tex]
So
[tex]p = \sqrt{ 2 * (1.67*10^{-27} ) * 225 * 1.6*10^{-19}}[/tex]
=> [tex]p = 3.4676 *10^{-22} \ kg \cdot m/ s[/tex]
So the de-Broglie wavelength isis mathematically represented as
[tex]\lambda = \frac{h}{p}[/tex]
Here h is the Planck's constant with value
[tex]h = 6.626 *10^{-34} \ J\cdot s[/tex]
=> [tex]\lambda = \frac{6.626 *10^{-34}}{3.4676 *10^{-22} }[/tex]
=>[tex]\lambda = 1.9109 *10^{-12} \ m[/tex]
