Question

If is the de-broglie wavelength for a proton accelerated through a potential difference of 100 v, the de-broglie wavelength for alpha-particle accelerated through the same potential difference is

Answer 1

wavelength=√(150/V) in Angstrom
V is the potential difference

Answer 2

${\huge{\underline{\underline{\mathcal{\red{♡ANSWER♡}}}}}}$

• Acceleration potential ${V=100 v}$ .The de Broglie wavelength $\lambda{is}$
• $\lambda{h/p}$=$\frac{1.227}{\sqrt{V}}$ nm
• $\lambda$=$\frac{1.227}{\sqrt{100}}$ nm=${0.124nm}$.
• The de Broglie wavelength associated with an electron in this case is of the order of X-ray wavelengths.