Essay About De Broglie Wavelength Of An Electron And Kinetic Energy

The De Broglie Wavelength
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PHYSICS 4, CHAPTER 3ADDITIONAL PROBLEMSWhat 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 pmANSWER: 112 pmThe 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^-31E=4.3ueVANSWER: 4.3 μeV3. 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/sFrom uncertainty principle: deltax*deltap>=nIf we want to have minimum uncertainty: deltax*deltap=nDeltap=0.5%*p=9.35*10^-27 kgm/sDeltax=n/deltap=(h/2pi)/deltap=(6.625*10^-34/2pi)/(9.35*10^-27)=11nmANSWER: 11 nmThe 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>=nFor least uncertainty: deltax*deltap=n => deltap=(h/2pi)/deltax=(6.625*10^-34/2pi)/(50*10^-12)= ANSWER: 2.1 × 10-24 kg.m/s5. An electron is confined to a one-dimensional, infinitely deep potential energy well of width L = 100 pm. What is the smallest amount of energy the electron can have?E=h^2/(8ma^2)*n^2Where h=6.625*10^-34; m=9.1*10^-31; a=100*10^-12; n=1E1=6.02*10^-18 J= 37.7eV        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 eVThe energy must be transferred to the electron is: deltaE=E3-E1= 301eVIf 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 byde-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