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Chapter#03: Introduction to Quantum Mechanics

Team Quanta gladly presents all possible short questions of Modern Physics & Electronics’ Chapter#03: Introduction to Quantum Mechanics.

Q.6.1 what is important of Schrodinger wave equation in physics?

Answer: Schrodinger equation is second ordered differential equation, which governs the variation in space and time of wave function Ψ for a wide range of problems. As we obtain solutions to problems in classical mechanics by manipulating Newton’s laws of motion, similarly we obtain solutions to atomic problems  by solving Schrodinger wave equation. It is right to say that Schrodinger wave equation is a machine in which potential energy function must be supplied as input, and output consists of the wave functions, quantum numbers and energy Eigen values, that characterize the quantum behavior of system.

 Q.6.2 what is meant by operator in quantum mechanics? Give an example.

Answer: Operator is a mathematical entity which when operates on a function, changes it. For example, consider equation

    \[\frac{d}{dx}\left(\sin{4x}\right)=4\cos{4x}\]

Here \frac{d}{dx} is an operator \hat{A} and sin 4x is called operand.

Q.6.3 what is meant by Eigen function? Is  an Eigen function for operator ?

Answer: If an operator  operates on a function Ψ such that

    \[\hat{A}\ \mathrm{\Psi}=a\mathrm{\Psi}\]

Then Ψ is called Eigen function for operator  corresponding to Eigen value ‘a’. Since \frac{\partial}{\partial x}\ e^{2x}=2e^{2x}.

So e^{2x}  an Eigen function for operator  .

Q.6.4 what is minimum allowed energy of electron and proton confined to a deep well of width equal to diameter of atomic nucleus? On this basis, should you expect to find electrons inside nucleus?

Answer: Energy of particle confined to a deep well of length L is given by,

    \[E=n^2\left(\frac{h^2}{8mL^2}\right)\]

Energy is minimum for n = 1,

    \[E=1^1\left(\frac{h^2}{8mL^2}\right)=\ \frac{\left(hc\right)^2}{8\left(mc^2\ \right)L^2}\ =\frac{\left(1240eV-nm\right)^2}{8\left(mc^2\right)L^2}=\frac{\left(1240MeV-fm\right)^2}{8\left(mc^2\right)\left(14fm\right)^2}=\ \frac{\left(88.57MeV\right)^2}{8\left(mc^2\right)}\]

For electron,

    \[E=\frac{\left(88.57MeV\right)^2}{8\left(0.511MeV\right)}=1919MeV\]

Energy of electron is larger than binding energies of particles inside the nucleus, this justifies that electron cannot exist inside the nucleus.

    \[E=\frac{\left(88.57MeV\right)^2}{8\left(938MeV\right)}=1.04MeV\]

Energy for proton is of order of binding energies of particles inside the nucleus; the justifies existence of proton inside the nucleus.

Q.6.5 where is maximum probability density of a particle trapped in an infinitely well of length L in nth state?

Answer: Probability density of a particle trapped in an infinitely well of length L in nth state is,

    \[P=\mathrm{\Psi}^\ast\left(x\right)\ \mathrm{\Psi}\ \ \left(x\right)=\left(\sqrt{\frac{2}{L}}\ sin\frac{n\pi}{L}\ x\right)\left(\sqrt{\frac{2}{L}}\sin{\frac{n\pi}{L}\ x}\right)=\frac{2}{L}\ {sin}^2\frac{n\pi}{L}x\]

It will be maximum when

    \[{sin}^2\frac{n\pi}{L}x=1\ \ \ \ \ \ \ \ \Rightarrow\frac{n\pi}{L}x=\left(2k+1\right)\frac{\pi}{2}\ \ \ \ \ \ \ \ \ \ \ \Rightarrow x=\left(2k+1\right)\frac{L}{2n}\]

Q.6.6 where is minimum probability density of a particle trapped in an infinitely well of length L in nth state?

Answer: Probability density of a particle trapped in an infinitely well of length L in nth state is,

    \[P=\mathrm{\Psi}^\ast\left(x\right)\ \mathrm{\Psi}\ \ \left(x\right)=\left(\sqrt{\frac{2}{L}}\ sin\frac{n\pi}{L}\ x\right)\left(\sqrt{\frac{2}{L}}\sin{\frac{n\pi}{L}\ x}\right)=\frac{2}{L}\ {sin}^2\frac{n\pi}{L}x\]

It will be minimum, when

    \[{sin}^2\frac{n\pi}{L}x=0\ \ \ \ \ \ \ \ \ \ \Rightarrow\frac{n\pi}{L}x=\pm k\pi\ \ \ \ \ \ \ \ \ \ \ \Rightarrow X=k\frac{L}{n}\]

Q.6.7 what must be width of an infinite well such that a trapped electron in state n = 6 has an energy of 4.70eV?

Answer: Energy of particle confined to a deep well of length L is given by,

    \[E=n^2\left(\frac{h^2}{8mL^2}\right)\ \Rightarrow L^2=\frac{n^2h^2}{8mE}\ \Rightarrow L=\frac{nh}{\sqrt{8mE}}=\frac{6\times6.63\times{10}^{-34}\ Js}{\sqrt{8\times9.1\times{10}^{-31}kg\times4.70\times1.6\times{10}^{-19}\ J}}\]

Q.6.8 Find fractional difference between two adjacent energy levels of a particle confined in one dimensional well of infinite depth. Hence obtain condition of classical mechanics.

Answer: The fractional difference between two adjacent energy levels of a particle confined in one dimensional well of infinite depth is,

    \[\frac{\Delta E}{E}=\frac{\left(n+1\right)^2\left(\frac{h^2}{8mL^2}\right)-n^2\left(\frac{h^2}{8mL^2}\right)}{n^2\left(\frac{h^2}{8mL^2}\right)}\ =\frac{\left(n+1\right)^2-n^2}{n^2}=\frac{n^2+2n+1-n^2}{n^2}=\frac{2n+1}{n^2}\]

For classical system n\rightarrow\infty,(\Delta E/E)\rightarrow0  i.e. system behaves as if it is continuous.

Q.6.9 why are the wave functions Ψ(x) = not physically possible for all values of x?

Answer: This function is not acceptable because it becomes infinite when

Q.6.10 Show that \Psi\left(x\right)=Bcoskx is a solution to Schrodinger time independent wave equation (V(x)=0) if k=\sqrt{\frac{2mE}{\hbar^2}}. Explain why this cannot be an acceptable wave function for a particle in box of length L, no matter what the value of k.

Answer: Time independent Schrodinger wave equation with (V(x) = o) is given by,

    \[-\frac{\hbar}{2m}\frac{d^2\mathrm{\Psi}\left(x\right)}{dx^2}=E\mathrm{\Psi}\left(x\right)\\]

Putting \mathrm{\Psi}\left(x\right)=Bcos\ kx,\ we\ obtain

    \[-\frac{\hbar}{2m}\frac{d^2}{{dx}^2}\left(B\cos{kx}\right)=EB\cos{kx\ \ \ \ \ \ \ \ \ \ \ \ \ }\Rightarrow-\frac{\hbar}{2m} \frac{d}{dx}\left(-Bk\sin{kx}\right)=EB\cos{kx}\]

    \[\Rightarrow \frac{\hbar^2k^2}{2m}=E \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow k^2=\frac{2mE}{\hbar} \ \ \ \ \ \ \ \ \ \ \ \ \ \Rightarrow k= \sqrt{\frac{2mE}{\hbar}}\]

Hence  is a solution to Schrodinger time independent wave equation (V\left(x\right)=0\ \ if\ k= \sqrt{\frac{2mE}{\hbar}} . The wave function for a particle in a box of length L must satisfy boundary conditions:

  1. \mathrm{\Psi}\left(x\right)=0\ at\ x=0
  2. \mathrm{\Psi}\left(x\right)=0\ at\ x=L

Here \mathrm{\Psi}\left(0\right)=B\ cos0=B\neq0 hence \mathrm{\Psi}\left(x\right)=B\cos{kx}  is not an acceptable wave function because it does not satisfy the required boundary condition, even though it is a solution to Schrodinger wave equation.

Q.6.12 For a quantum particle in a box, the probability density at certain points is zero as seen in figure. Does this value imply that the particle cannot move across these points?

Answer: No, the motion of the quantum particle does not consist of moving through successive points and the particle has no definite position. It can sometimes be found on one side of a node and sometimes on the other side, but never at the node itself.

Q.6.13 In quantum mechanics, it is possible for the energy E of a particle to be less than the potential energy, but classically this condition is not possible. Why?

Answer: According to classical mechanics, the kinetic energy is never negative i.e.E\geq V  In quantum mechanics, particles are represented by wave functions, the Schrodinger equation shows that there is a nonzero probability that a particle can tunnel through a barrier- a region in which E<V.

Q.6.14 what is the relationship between ground-state energy and the uncertainty principle?

Answer: Consider a particle in one dimensional box. If minimum energy of particle were zero, then the particle could have zero momentum. At the same time, the uncertainty in its position would not be infinite, but equal to the width of the box. In such a case, the uncertainty product  would be zero, violating the uncertainty principle. Thus the minimum energy of the particle is not zero.

Q.6.15 If a particle in a box is in nth energy level, what is average value of its x-component of momentum?

Answer: Stationary wave functions for a particle in a box shows that they are resultant of waves propagating in opposite directions. One wave has momentum in +x-direction and other has an equal momentum in –x-direction, giving total momentum zero in x-direction i.e. .

Review Questions

R.6.1 How is the Schrodinger equation useful in describing quantum phenomena?

Answer: The motion of macroscopic objects can be determined by Newton’s law whereas Schrodinger equation is used to determine the wave function of a particle of small mass. In particular, the states of atomic electrons are confined wave states whose wave functions are solutions to the Schrodinger equation. Anything that we can know about a particle comes from its wave function.

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