LinkedIn Insight Summer 2016 - Q.5 - Grad Plus

Summer 2016 – Q.5

5. (a) Deduce Schrodinger’s time independent wave equation. [6M]

Schrodinger’s wave equation determines the motion of atomic particle. Consider a microparticle in motion which is associated with wave function Ψ. The Ψ represents the field of the wave, say. The classical wave equation is of the form,
$latex \frac{\partial ^{2}y}{\partial x^{2}}=\frac{1}{v^{2}} \frac{\partial ^{2}y}{\partial t^{2}}$
A solution of this equation is, $latex y\left ( x,t \right )=Ae^{-i\left ( kx-wt \right )}$
For a microparticle the w and k can be replaced with E and P using Einstein and de-Broglie relations as,
E= ћω and p= ћk,
Also replacing y(x,t) by Ψ(x,y), we may write
$latex \psi\left(x,t\right)=Ae^{-i\left(Et-px\right)/\hslash}$
Differentiating with respect to t, we get
$latex \frac{\partial\psi}{\partial t}=\frac {i}{\hslash}E\psi$
Differentiating with respect to t we get
$latex \frac{\partial^2\psi}{\partial x^2}=\frac{p^2}{\hslash^2}E\psi$
For a free particle we have E=p2/2m and in case of a particle moving in force field characterized by potential energy V, we have p2/2m=E-V.
Multiplying above equation by ψ, $latex \frac{p^{2}}{2m}\psi =E\psi -V\psi $
Substituting for Eψ and p2ψ, and rearranging we get
$latex -\frac{\hslash^2}{2m}\frac{\partial^2\psi}{\partial x^2}+V\psi=i\hslash\frac{\partial\psi}{\partial t}$
Which is time dependent Schrodinger’s wave equation, where $latex i\hslash\frac\partial{\partial t}=E,$ the energy operator.
$latex -\frac{\hslash^2}{2m}\frac{\partial^2\psi}{\partial x^2}+\hslash\psi=E\psi$

Which is time independent Schrodinger’s wave equation.

(b) State and explain Heisenberg’s uncertainty principle. [4M]

Statement: “It is impossible to determine the exact position and exact momentum both of a particle simultaneously with unlimited accuracy.”

It means that, if the position of the particle is determined accurately then there is uncertainty in the determination of momentum and if the momentum of the particle is determined accurately then there is uncertainty in the determination of its position.

Mathematically, “The product of uncertainties in the simultaneous measurement of position and momentum of a microscopic particle is always of the order of Planck’s constant or greater than or equal to h/2π”.

Thus mathematically,

Δx Δpx ≥ ђ [ ђ = h/2 π ]
Δx Δpx ≥ h/2π

This is called as Heisenberg’s uncertainty relation. Where, Δx = Uncertainty in the measurement of position, ΔpX = uncertainty in the measurement of momentum.
If Δx is small i.e. position is determined more accurately then ΔpX will be large i.e. momentum is determined less accurately and vice-versa.
The uncertainty principle also restricts the precision in the determination of particle energy ‘E’ at the time ‘t’. The uncertainty relation for energy is given by,

ΔE Δt ≥ ђ

Where, ΔE = Uncertainty in the measurement of energy, Δt = uncertainty in time.
The uncertainty relation for the angular momentum ( l ) of a particle in motion and the angular position (θ) is given by,

Δl Δθ ≥ ђ

Where, Δl = Uncertainty in the measurement of angular momentum, Δθ= uncertainty in the measurement of angular position.

(c) Calculate de Broglie wavelength for a proton moving with velocity 1 percent of the velocity of light. [3M]
Velocity of proton, $latex V=1\%\;of\;C=\frac1{100}\times3\times10^8=3\times10^6\;m/s.$
We have,

$latex \begin{array}{l}\lambda=\frac h{mv}\\\\\;\;\;=\frac{6.63\times10^{-34}}{1.67\times10^{-27}\times3\times10^6}\\\\\;\;\;=1.32\times10^{-13}m.\end{array}$

Scroll to Top
error: Alert: Content selection is disabled!!