λ = = …….(1) = …….(2) 
= = λ 
(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 1034Js/ 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 1034Js/ 4.50 kg m/s =1.47 ×10–34 m

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 m = 0.5kg, h = 6.626 × 1034 m2 kg / s Δp =p×1×10−6 As we know that, Δp = 105 Heisenberg Uncertainty principle formula is given as, ∆x∆p ∆x ∆x ∆x =0.527 x 1029 m 
∆x∆px = ℏ/2 or Thus ∆px= ℏ /2∆x ∆px= ℏ /2a/2 = ℏ /a Uncertainty in momentum of particle along xaxis is ∆px= ℏ /a 
K.E. = = = 
E = = =8.33 x 1015J =eV52.05 keV 
∆ 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 10101 units, (for quantum statistics) being .

Thus, uncertainly in energy is ∆ E
Or ∆ E J
Or ∆ E 5.3 x 1027 J
Frequency of light is uncertain by
∆ ν = = Hz 0.8 x 107 Hz As a result, the radiation from an excited atom does not have the noted precise frequency new ν  ∆ ν and ν + ∆ ν. 
E= K+V =mv2+V(x) = +V(x) ……….. (1) 
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 timedependent Schrodinger equation.

=  + V(x) ψ(x,t) ……….. (8) 
=  + 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 JouleSec. 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 
m=10 gm. h=6.6×10–27 ergsec Δv= 5.25×10–26 cm Δx=? Δx.Δv ≥h/4πm Δx≥h/4πmΔv ≥6.6×1034/4×3.14×10×5.25×10–26 ≥ 0.10×10–2m ≥ 1×10–3 cm 
ψ(x, y, z, t) = a + ib and its complex conjugate as ψ*(x, y, z, t) = a – ib. The product of wave function and its complex conjugate is ψ(x, y, z, t)ψ*(x, y, z, t) = (a + ib) (a – ib) = a2 + b2 a2 + b2 is a real quantity. 
dx =1 
V= 0 for 0 ………. (1) V= for x <0, x>L ………. (2) 
Figure : Particle in deep potential well 
ψ + E ψ =0 …………….(2) Substituting E = k2 …………….(3)
writing the SWE for 1D 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 
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.
U=U0 x <0 (1) U= 0 0x L (2) U=U0 L < x. (3) 
We want to solve Schroedinger’s Equation for this potential to get the wave functions and allowed energies for E < U0. We will refer to the three regions as regions 0, 1, and 2 with associated wave functions ψ0,ψ1,ψ2. Figure 8: Finite Square Well Potential Energy The timeinvariant, nonrelativistic 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 implicitly 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,

Figure : Wave function in Finite well 