λ =   =    …….(1)    =     …….(2)

(a)For the electron:

 Mass m = 9.11×10–31 kg, speed v = 5.4×106 m/s.

Then, momentum

p = m v = 9.11×10–31 kg × 5.4 × 106 (m/s)

p = 4.92 × 10–24 kg m/s

de Broglie wavelength, λ = h/p = 6.63 x 10-34Js/ 4.92 × 10–24 kg m/s

λ= 0.135 nm

(b)For the ball:

Mass m’ = 0.150 kg,

Speed v ’= 30.0 m/s.

Then momentum p’ = m’ v’= 0.150 (kg) × 30.0 (m/s)

p’= 4.50 kg m/s

de Broglie wavelength λ’ = h/p’ =6.63 x 10-34Js/ 4.50 kg m/s =1.47 ×10–34 m

The de Broglie wavelength of electron is comparable with X-ray wavelengths. However, for the ball it is about 10–19 times the size of the proton, quite beyond experimental measurement.

Accelerating potential V = 100 V. The de Broglie wavelength λ is

λ= h /p = 1 227/ nm

λ.1 227/  nm = 0.123 nm

The de Broglie wavelength associated with an electron in this case is of the order of x ray wavelengths.

Given
v = 20m/s,

m = 0.5kg,

h = 6.626 × 10-34 m2 kg / s

Δp =p×1×10−6

As we know that,
P = m×v = 0.5×20 = 10kgm/s
Δp = 10×1×10−6

Δp = 10-5

Heisenberg Uncertainty principle formula is given as,

∆x∆p

∆x

∆x

∆x =0.527 x 10-29 m

∆x ∆p =h/2π

Thus ∆p=h/2π∆x

Or ∆p=6.62 x10-34/2 x 3.14 x 10-14

Or ∆p=1.05 x 10-20 kg m/ sec

E =

∆x ∆p =h/2π

Thus ∆p=h/2π∆x

Or ∆p=6.62 x10-34/2 x 3.14 x 10-14

Or ∆p=1.05 x 10-20 kg m/ sec

If this is p the uncertainty in the momentum of electron then the momentum of electron should be at least of this order that is p=1.05*10-20 kg m/sec.

An electron having this much high momentum must have a velocity comparable to the velocity of light. Thus, its energy should be calculated by the following relativistic formula

E =

∆x = a/2 (half the diameter).

According to uncertainty principle

∆x∆px = ℏ/2

or

Thus    ∆px= ℏ /2∆x

∆px= ℏ /2a/2 = ℏ /a

Uncertainty in momentum of particle along x-axis is

∆px= ℏ /a

Assuming the momentum of particle to be at least equal to uncertainty in it, the lowest possible value of K.E. of particle is given by

K.E. = = =

E = = =8.33 x 10-15J =eV 

52.05 keV

∆ x ∆ px = ;

Similarly  ∆ y ∆ py = ,

And   ∆ z ∆ pz = ,

Multiplying these three equations, we get

 ∆ x ∆ y ∆ z ∆ px  ∆ py ∆ pz = ( )3  in the units (J3 S3).

The above product is called the volume of elementary cell in phase space.

So, volume of an elementary cell in phase space 10-101 units, (for quantum statistics) being .

∆ E

Or  ∆ E J

Or  ∆ E 5.3 x 10-27 J

Frequency of light is uncertain by

 ∆ ν = = Hz 0.8 x 107 Hz

E= K+V =mv2+V(x) = +V(x)   ……….. (1)

By de Broglie’s relation we know that all particles can be represented as waves with frequency ω and wave number k, and that E= ℏω and p= ℏk.

Using this equation (1) for the energy will become

ℏω = + V(x)    ……….. (2)

A wave with frequency ω and wave number k can be written as usual as

ψ(x, t) =Aei(kx−ωt)     ……….. (3)

the above equation is for one dimensional and for three dimensional we can write it as

ψ(r, t) =Aei(k·r−ωt)    ……….. (4)

But here we will stick to one dimension only.

=−iωψ   ⇒   ωψ=   ……….. (5)   

=−k2ψ  ⇒   k2ψ = -    ……….. (6)

If we multiply the energy equation in Eq. (2) by ψ, and using(5) and (6) , we obtain

ℏ(ωψ)  = ψ+ V(x) ψ  ⇒   = - + V(x) ψ  ……….. (7)

This is the time-dependent Schrodinger equation.

If we put the x and t in above equation then equation (7)  takes the form as given below

= - + V(x) ψ(x,t)    ……….. (8)

In 3-D, the x dependence turns into dependence on all three coordinates (x, y, z) and the  term becomes ∇2ψ.

The term |ψ(x)|2 gives the probability of finding the particle at position x.

Let us again take it as simply a mathematical equation, then it’s just another wave equation. However We already know the solution as we used this function ψ(x, t) =Aei(kx−ωt) to produce Equations (5), (6) and (7)

But let’s pretend that we don’t know this, and let’s solve the Schrodinger equation as if we were given to us. As always, we will guess an exponential solution by looking at exponential behaviour in the time coordinate, our guess is ψ(x, t) =e−iωtf(x) putting this into Equation (7) and cancelling the e−iωt yields

  = - + V(x) f(x)   ……….. (9)

We already know that E=. However ψ(x, t) is general convention to also use the letter ψ to denote the spatial part. So we will now replace f(x) with ψ(x)

   Eψ = - + V(x) ψ    ……….. (10)

Given     m= 9×10–31

ΔV= 5.7×105 m/sec                 

Δx=?

h= 6.6×10–34 Joule-Sec.

Δx.Δv ≥h/4πm

Δx≥h/4πmΔv

≥6.6×10–34/9×3.14×9.1×10–31×5.7×105

≥ 0.010×10–8

≥ 1×10–10m

Given

m=0.15 kg.     

h=6.6×10-34 Joule-Sec.

Δx = 10 –10 m

Δv=?

Δx.Δv ≥h/4πm

Δv≥h/4πmΔx

≥6.6×10-34/4×3.14×0.15×10–10

≥ 3.50×10–24m

m=10 gm.                    h=6.6×10–27 erg-sec

Δv= 5.25×10–26 cm

Δx=?

Δx.Δv ≥h/4πm

Δx≥h/4πmΔv

≥6.6×10-34/4×3.14×10×5.25×10–26

≥ 0.10×10–2m

≥ 1×10–3 cm

Let us consider a particle of mass ‘m’ in a deep well restricted to move in a one dimension (say x). Let us assume that the particle is free inside the well except during collision with walls from which it rebounds elastically.

The potential function is expressed as

V= 0 for 0      ………. (1)

V= for x <0, x>L     ………. (2)

Figure : Particle in deep potential well

The probability of finding the particle outside the well is zero (i.e. Ѱ =0) 

 Inside the well, the Schrödinger wave equation is written as

ψ + E ψ =0    …………….(2)    

Substituting   E = k2                                       …………….(3)

writing the SWE for 1-D we get

+ k2 ψ =0       …………….(4)

The general equation of above equation may be expressed as

ψ = Asin (kx + ϕ)      …………….(5)

Where A and ϕ  are constants to be determined by boundary conditions

Condition I: We have ψ = 0 at x = 0, therefore from equation

0 = A sinϕ   

As A then sinϕ =0  or  ϕ=0    …………….(6)

Condition II: Further ψ = 0 at x = L, and ϕ=0 , therefore  from equation (5)

0 = Asin kL

As A then sinkL =0  or  kL=nπ

k =       …………….(7)

where n= 1,2,3,4………

Substituting the value of k from (7) to (3)

)2 = E

 This gives energy of level  

En =    n=1,2,3,4…so on    …………….(8)

From equation En is the energy value (Eigen Value) of the particle in a well.

It is clear that the energy values of the particle in well are discrete not continuous.

Using (6) and (7) equation (5) becomes, the corresponding wave functions will be

ψ = ψn = Asin   …………….(9)

The probability density

|ψ(x,t)|2 = ψ ψ*

|ψ(x,t)|2 = A2sin2   …………….(10)

The probability density is zero at x = 0 and x = L. since the particle is always within the well

    …………….(11)

=1

A =

Substituting A in equation (9) we get

ψ = ψn = sin    n=1,2,3,4…..    …………….(12)

The above equation (12) is normalized wave function  or Eigen function belonging to energy value En

Figure: Wave function for Particle

Therefore, according to uncertainty principle it is difficult to assign a position to the electron.

=-iω (x, t)  and =- (π2/a2)  (x, t) the Schrödinger equation yields

Hence V(x,t) is time independent and given by

(b) The probability of finding the particle in the interval a/4 x 3a/4.can be obtained from

Substituting these values in the time dependendent Schrödinger equation, we have

U=U0     x <0      (1)

U= 0  0x L      (2)

U=U0  L < x.       (3)

Figure 8: Finite Square Well Potential Energy

The time-invariant, non-relativistic Schroedinger’s equation is

         (4)

that can be rearranged to give

         (5)

It is convenient to define two new variables (both positive), one for regions 0 and 2, and one for region 1—they are wavenumbers:

         (6)

            (7)

and Schrödinger’s equation becomes

In regions 0 and 2 the general solution is a linear combination of exponentials with the same form, but with different constants, namely

In region 1 we have the same general solution that we had for the infinite square well,

Equations (10) to (12) have 7 unknowns—A,B,C,D,F,G and the energy E that is im-plicitly contained in the variables κ0,k1. Therefore we need to get 7 equations to be ableto solve for the unknowns.

We will first use the requirement that the wavefunction remain finite everywhere. Consider ψ2 as x→∞. For this to remain finite we must require G= 0. Similarly, as x→−∞, we require A= 0. Our solutions become

The next step is to require that the wavefunction and its first derivative be continuous everywhere, and in our case we look at the boundaries, x= 0 and x=L.

Hence ψ0=D exp(+κ0x).Take derivatives of the wave functions,

There remain 3 unknowns, D,F, and E. Finding them is a bit messier! Consider the boundary conditions at x=L,

We are not going much farther, but if we divided Equation (22) by Equation (23), we can see that the constants D and F cancel leaving us with one rather difficult equation to solve for energy E (remember this is implicitly included in the values of k1and κ0.)

There is one remaining condition, normalization, that for this problem is

Even without solving the entire problem we can make some conclusions about the wave-function and the allowed energy levels. Recall that for an infinite square well potential of width L the allowed energies are quantized and

Here is Equation(22)/Equation(23)

Now put in the values ofκ0andk1from Equations (6) and (7) and do some algebra to get

This equation is a single equation in a single unknown, E < U0. Once we have the details of our particle (its mass) and the potential energy (depth U0 and width L), we can solve it. There is no analytic solution, only a numerical one.

From equation (25) With n being any positive integer. Outside the well the wavefunction is 0. We are certain that the particle is somewhere inside the box, so ∆x ∞ =L.

With the finite well, the wavefunction is not zero outside the well,

Figure 9: Wave function in Finite well

So ∆x finite> L, hence from the uncertainty principle, ∆pfinite x<∆p∞x. This suggests that the average value of momentum is less for the finite well, and therefore that the kinetic energy inside the well is less for the finite well than for the infinite well. Indeed this is borne out with detailed analysis. In addition, the number of allowed energy levels is finite, and there is a possibility that a well may be sufficiently narrow or sufficiently shallow that no energy levels are allowed. Also note that the non-zero wave functions in regions 0 and 2 mean that there is a non-zero probability of finding the particle in a region that is classically forbidden, a region where the total energy is less than the potential energy so that the kinetic energy is negative.

u (x, t) =A sin(ωt - kx)    …….(1)

u’(x, t) =A sin[(ω+∆ω)t-(k+∆k)x]   …….(2)

Where ω  and (ω+Δω)  are angular frequencies and k and (k +Δk) are propagation constants

The superposition of two waves is of the form

ψ(x,t) = u +u’ = A sin(ωt - kx)sin[(ω+∆ω)t - (k+∆k)x]      …….(3)

As Δ ω and Δk are small therefore (ω+Δω) and (k + Δk) k, equation (3) reduces to

ψ(x,t) = 2Acos[ t -x] sin(ωt - kx)    .…….(4)

This equation represents a vibration of amplitude

2Acos[ t -x]       …….(5)

The phase of the resultant wave moves with the velocity known as phase velocity

vp=          ……..(6)

And the amplitude moves with the velocity known as group velocity

vg = = 

vg =          ….….(7)