QUANTUM MECHANICS PROBLEMS

 

Q/1. For an electron in a one-dimensional infinite potential well of width 1 Å, calculate (i) the separation between the two lowest energy levels; (ii) the frequency and wavelength of the photon corresponding to a transition between these two levels ?  

Ans : (i)   E_n = π²ђ²n²/2ma²    ;   2a = 1Å

            E₂ – E₁ = (4 -1)π²ђ²/2ma²   

                       = 3*π²(1.055-10^-34 )²/8*9.1*10^-31 *10^-20  

                       = 1.812*10^-17 J = 113.27 eV  

   (ii) hν = E₂ – E₁ = 1.812*10^-17 J

                 hc/λ = 1.812*10^-17 J

                 λ= 6.62*10^-34 *3*10⁸/1.812*10^-17 =1.1*10^-8 m

 

Q/2. Calculate the expectation values of position <x> and of the momentum <p_x> of the particle trapped in the one-dimensional box  in the potential V(x) = 0 for 0≤ x≤ a  and V(x) = ∞ otherwise?

 Ans : <x> =(2/a) ₀∫^a  sin(nπx/a) *x* sin(nπx/a) dx

                   =(2/a) ₀∫^a x sin²(nπx/a) dx =(1/a) ₀∫^a x*[1- cos(2nπx/a)] dx

                  =(1/a) ₀∫^a x dx - (1/a) ₀∫^a  cos(2nπx/a) dx

        <x> = a/2

          <p_x> = (2/a) ₀∫^a sin(nπx/a) {-iђ(d/dx)}* sin(nπx/a) dx

                  =-iђ(2nπ/a²) ₀∫^a sin(nπx/a) *cos(nπx/a) dx

                 =-iђ(nπ/a²) ₀∫^a sin(2nπx/a) dx = 0

 

 Q/3. An electron in a one-dimensional infinite potential well, defined by V(x) = 0 for 0≤ x≤ a  and V(x) = ∞ otherwise, goes from the n = 4 to the n = 2 level. The frequency of the emitted photon is 3.43 *10¹⁴ Hz Find the width of the box ?

Ans :   E_n = π²ђ²n²/2ma²    ;   

            E₄ – E₂ = (16 -4)π²ђ²/2ma² = hν

           a² = 3h/8mν = 3*6.626*10^-34/(8*9.1*10^-31 *3.43*10¹⁴)

                 =   79.6*10^-20 m²

           a = 8.92*10^-10m

 

 Q/4. A particle of mass m trapped in the potential V(x) = 0 for 0≤ x≤ a  and V(x) = ∞ otherwise, Evaluate the probability of finding the trapped particle between x = 0 and x = a/n when it is in the nth state?

Ans :  Wave function ψ(x) = √(2/a)  sin(nπx/a)

          Probability density P(x) = (2/a)  sin²(nπx/a)

           Required probability P =₀∫^a/n P(x)dx =  ₀∫^a/n (2/a)  sin²(nπx/a) dx

                                  P  =(1/a) ₀∫^a/n[1- cos(2nπx/a)] dx

                                  =1/n

 

Q/5. The wave function of a particle confined in a box of length a is

                          ψ(x) = √(2/a)  sin(nπx/a)  ;           0≤ x ≤ a

 Calculate the probability of finding the particle in the region 0 < x < a/2 ?

Ans : The required probability P = ₀∫^a/2 P(x)dx =(2/a) ₀∫^a/2 sin²(πx/a) dx

                =(1/a) ₀∫^a/2[1- cos(2πx/a)] dx

              =(1/a) ₀∫^a/2 dx - (1/a) ₀∫^a/2 cos(2πx/a) dx                    

                 = ½                      

Q/6. An electron is in the ground state of a one-dimensional infinite square well with a = 10^–10 m. Compute the force that the electron exerts on the wall during an impact on either wall ?

 Ans :  The force on the wall ;    F = -(dE_n/da)

              The energy of the ground state ; E₁= π²ђ²/2ma²    

          Hence the force on the wall F = -(dE_n/da)|_{a= 10}^-10   = π²ђ²/2ma³

                           = π²(1.054*10^-34)²/{9.1*10^-31*(10^-10)³}

                         =1.21*10^-7N