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COURSE-B TECH
SUBJECT- CHEMISTRY
MOST IMPORTANT DERIVATIONS FOR
EXAMS, SCHRODINGER WAVE EQUATION
AND ENERGY OF PARTICLE IN ONE
DIMENSION(PARTICLE IN A BOX)
Derivation of Schrödinger's Equation:-
So, when we talk about the wave function, which is represented as
Ψ(x,t), it basically tells us the probability of finding an electron at a specific spot (position 'x') and at a certain time ('t')^1.
Now, for a wave function to be valid, it needs to have a few key characteristics:
● It should always be a single value. ● It has to be continuous. ● And, importantly, it must be well-defined.
To derive the core of it, the Hamiltonian operator
H^ acting on the wave function Ψ(x,t) gives us the energy E multiplied by the wave function Ψ(x,t)^4.
The general wave equation looks like this:
∂x2∂2Ψ(x,t)=V2VΨ∂t2∂2Ψ(x,t) 5
Here, 'V' is velocity, 'x' is position, and 't' is time.
We can separate the wave function into a spatial part and a time-dependent part:
Ψ(x,t)=Ψ(x)Ψ(t) 7
More specifically, the time-dependent part often involves a cosine function:
Ψ(x,t):Ψ(x)cosωt 8
And
ω is the angular frequency, which is 2πν^9.
When we apply the second derivative with respect to x on the wave function and the second derivative with respect to t, we get:
∂x2∂2[Ψ(x)cosωt]=V21∂t2∂2[Ψ(x)cosωt] 10
This simplifies to:
cosωt∂x2∂2Ψ(x)=V21Ψ(x)∂t2∂2(cosωt) 11
Which further reduces to:
cosωt∂x2∂2Ψ(x)=V21Ψ(x)[−ω2cosωt] 12
Substituting 'p', we get:
λ=2m(E−U)h
To find V2ω2:
V2ω2=(V2πν)2=V24π2ν2=(λc)2=(λν)
This can also be expressed as:
=hV2m(E−U)
Substituting the value of ω2 into V2ω2:
V2ω2=h24π2[2m(E−U)]
Now, using
ℏ =2πh and ℏ 2=4π2h2, the equation can be written as:
ℏ 22m(E−U)
Substituting this into the wave equation, we get the time-independent Schrödinger Equation:
∂x2∂2Ψ(x)+ ℏ 22m(E−U)Ψ(x)=
Expanding it:
∂x2∂2Ψ(x)+ ℏ 22mEΨ(x)− ℏ 22mUΨ(x)=0 2828
Rearranging:
∂x2∂2Ψ(x)−ℏ22mUΨ(x)=ℏ2−2mEΨ(x) 2929
Multiplying the whole equation by 2mℏ 2 and changing the sign, we arrive at the 1D Schrödinger Equation:
−2mℏ 2 ∂x2∂2Ψ(x)+UΨ(x)=EΨ(x) 30
This is often written in a more compact form using the Hamiltonian operator:
H^Ψ(x)=EΨ(x) 31
Where the Hamiltonian operator for one dimension is:
H^=[2m− ℏ 2 ∂x2∂2+U]
For two dimensions (x,y), it would look like this:
Ψ(x)=AcosKx+BsinKx 39
Now, applying the boundary conditions:
- At x=0, Ψ(0)=0^40 :
AcosK(0)+BsinK(0)=0 41
A(1)+B(0)=0 42
This means A=0^43.
- Substituting A=0 back into our general solution, we get:
Ψ(x)=BsinKx 44
- Now, at x=L, Ψ(L)=0^45 :
BsinKL=0 46
Since B cannot be zero (otherwise there's no wave function), sinKL must be zero^47.
This implies KL=nπ, where n is an integer (0,1,2,...)^48.
So, K=Lnπ^49.
To relate this back to energy, let's look at the second derivative of Ψ(x):
∂x∂Ψ(x)=KBcosKx 50
∂x2∂2Ψ(x)=−K2BsinKx 51
Since Ψ(x)=BsinKx, we can write:
∂x2∂2Ψ(x)=−K2Ψ(x) 52
Rearranging this gives:
∂x2∂2Ψ(x)+K2Ψ(x)=0 53
Comparing this with the simplified Schrödinger equation (when U=0):
∂x2∂2Ψ(x)=ℏ2−2mEΨ(x) 54
Or,
∂x2∂2Ψ(x)+ℏ22mEΨ(x)=0 55
By comparing the two expressions for the second derivative of Ψ(x), we find:
K2=ℏ22mE 56
