The De Broglie Wavelength

PHYSICS 4, CHAPTER 3

ADDITIONAL PROBLEMS

1. What is the de Broglie wavelength of an electron with a kinetic energy of 120 eV?

E=1/2mv^2 =>v=căn(2*E/m)=căn(2*(120*1.6*10^-19)/(9.1*10^-31)=6.5*10^6(m/s)

• Lamda=h/p=h/(mv)=(6.625*10^-34)/(9.1*10^-31*6.5*10^6)=112 pm

ANSWER: 112 pm

1. The wavelength of the yellow spectral emission line of sodium is 590 nm. At what kinetic energy would an electron have that wavelength as its de Broglie wavelength?

P=mv=h/lamda=1.12*10^-27

E=1/2mv^2 =1/2*p^2/m for m=9.11*10^-31

• E=4.3ueV

ANSWER: 4.3 μeV

3. Assume that an electron is moving along an x axis and that you measure its speed to be 2.05 × 106 m/s, which can be known with a precision of 0.50%.

What is the minimum uncertainty (as allowed by the uncertainty principle in quantum theory) with which you can simultaneously measure the position of the electron along the x axis?

 P=mv=9.11*10^-31*2.05*10^-6=1.87*10^-24 kgm/s

From uncertainty principle: deltax*deltap>=n

If we want to have minimum uncertainty: deltax*deltap=n

• Deltap=0.5%*p=9.35*10^-27 kgm/s

Deltax=n/deltap=(h/2pi)/deltap=(6.625*10^-34/2pi)/(9.35*10^-27)=11nm

ANSWER: 11 nm

1. The uncertainty in the position of an electron along an x axis is given as 50 pm, which is about equal to the radius of a hydrogen atom. What is the least uncertainty in any simultaneous measurement of  the momentum of this electron?

Deltax*deltap>=n

For least uncertainty: deltax*deltap=n => deltap=(h/2pi)/deltax=(6.625*10^-34/2pi)/(50*10^-12)=

ANSWER: 2.1 × 10-24 kg.m/s

5. An electron is confined to a one-dimensional, infinitely deep potential energy well of width

L = 100 pm.

1. What is the smallest amount of energy the electron can have?

E=h^2/(8ma^2)*n^2

Where h=6.625*10^-34; m=9.1*10^-31; a=100*10^-12; n=1

E1=6.02*10^-18 J= 37.7eV        

1. How much energy must be transferred to the electron if it is to make a quantum jump from its ground state to its second excited state?

Second state: n=3=> E3=339 eV

The energy must be transferred to the electron is: deltaE=E3-E1= 301eV

1. If the electron gains the energy for the jump from energy level E1 to energy level E3 by absorbing light, what light wavelength is required?

  hf=deltaE=Ehigh-Elow

 f=c/lamda

=> lamda=h*c/deltaE=(6.625*10^-34)*3*10^8/301eV= 4.12nm

(d) Once the electron has been excited to the second excited state, what wavelengths of light can it emit by

de-excitation?

                The light is emitted, not absorved, the electron can jumps directly to the ground state by emitting light of wavelength

        Lamda=4.12*10^-9 m