Question

What is the relation between the wavelength, λ, and the momentum, p, of a particle? Roughly what kinetic energy should an electron have in order for its wavelength be of roughly an "atomic" distance scale? Re-express this as the number of volts needed to accelerate that electron from rest to this energy. Would you call this a high-voltage device?

Answer 1

λ = h/p

`E=2.41*10^(-17)Joules`

This energy is equivalent to E=150.44eV

Step-by-step explanation:

de Broglie Equation:

λ = h/p

h=6.63×10-34 m2 kg / s Planck's constant

so: p=h/λ

On the other hand, an atomic distance, an angstrom, is equal:
`10^(-10)m`, λ should be similar to this distance.

Kinetics Energy of a electron, m=9.11×10−31 kg

`E=(p^(2))/(2m_e)=(h^(2))/(2\lambda^(2)m_e) =((6.63*10^(-34))^(2))/(2*(10^(-10))^(2)*9.11*10^(-31))=2.41*10^(-17)Joules`

Number of volts needed to accelerate that electron from rest to this energy:

`E=Vq_(e)`

`V=E/q_(e)=2.41*10^(-17)/(1.602*10^(-19))=150.44Volts`

So, this energy is equivalent to E=150.44eV

This high-voltage device is called: particles accelerator