Question

What is the de-Broglie wave length associated with an electron accelerated through a potential difference of 100 volts?

Answer 1

According to Louis de Broglie's famous equation, Wave Nature (Wavelength) and Particle Nature (Momentum) are related by the following equation:

Wavelength = Planck's Constant/Momentum

But, Momentum and Kinetic Energy may be related as:

Momentum = sqrt(2 x Mass x Kinetic Energy) [ As K.E. = 0.5mv^2, and Momentum = mv]

But, Let us assume that is entire Kinetic Energy is used by the particle to overcome the Potential Difference. In that case,

K.E = Energy required by electron to cross given Potential Difference

= Charge of Electron x Potential Difference

= 1.6 x 10^(-19) C x 100 V

= 1.6 x 10^(-17) J

In addition to this, it should be known that massof an electron is 9.1 x 10^(-31) kg.

So, required momentum = sqrt(2 x 9.1 x 10^(-31) x 1.6 x 10^(-17)) kg m s^(-1)

= 5.39 x 10^(-24) kg m s^(-1)

So, plugging this value of momentum into de Broglie's equation,

Wavelength = Planck's Constant/Momentum

= 6.626 x 10^(-34) / 5.39 x 10^(-24) m

= 1.2293 x 10^(-10) m

= 1.2293 Å

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.