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IGNOU MPH-004 Solved Assignment 2024 | MSCPH | IGNOU

Solved By – Narendra Kr. Sharma – M.Sc (Mathematics Honors) – Delhi University

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IGNOU MPH-004 Assignment Question Paper 2024

mph-004-efbd489a-1e9c-48fe-a38c-14df4f364cd8

mph-004-efbd489a-1e9c-48fe-a38c-14df4f364cd8

PART A
  1. a) Calculate the average de Broglie wavelength of a nitrogen molecule at room temperature, given that the mass of nitrogen molecule is 4.65 × 10 26 k g 4.65 × 10 26 k g 4.65 xx10^(-26)kg4.65 \times 10^{-26} \mathrm{~kg}4.65×1026 kg.
    b) A particle of mass m m mmm is constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance L 0 L 0 L_(0)L_0L0. Using the uncertainty principle, determine the zero-point energy of the particle.
    c) The wavefunction for a particle is given by:
ψ ( x ) = { N ( L 2 x 2 ) L x L 0 elsewhere ψ ( x ) = N L 2 x 2 L x L 0 elsewhere psi(x)={[N(L^(2)-x^(2)),-L <= x <= L],[0,” elsewhere “]:}\psi(x)=\left\{\begin{array}{cc} N\left(L^2-x^2\right) & -L \leq x \leq L \\ 0 & \text { elsewhere } \end{array}\right.ψ(x)={N(L2x2)LxL0 elsewhere
Calculate the normalization constant N N NNN.
d) Show that [ L ^ x L ^ y , L ^ z ] = i ( L ^ x 2 L ^ y 2 ) L ^ x L ^ y , L ^ z = i L ^ x 2 L ^ y 2 [ hat(L)_(x) hat(L)_(y), hat(L)_(z)]=iℏ( hat(L)_(x)^(2)- hat(L)_(y)^(2))\left[\hat{L}_x \hat{L}_y, \hat{L}_z\right]=i \hbar\left(\hat{L}_x{ }^2-\hat{L}_y{ }^2\right)[L^xL^y,L^z]=i(L^x2L^y2)
2. a) Calculate the expectation values x ^ x ^ (: hat(x):)\langle\hat{x}\ranglex^ and x ^ 2 x ^ 2 (: hat(x)^(2):)\left\langle\hat{x}^2\right\ranglex^2 for the following odd parity state of a symmetric infinite potential well
ψ ( x ) = 1 a sin ( n π x 2 a ) ; n = 2 , 4 , 6 . ψ ( x ) = 1 a sin n π x 2 a ; n = 2 , 4 , 6 . psi(x)=(1)/(sqrta)sin((n pi x)/(2a));n=2,4,6dots.\psi(x)=\frac{1}{\sqrt{a}} \sin \left(\frac{n \pi x}{2 a}\right) ; n=2,4,6 \ldots .ψ(x)=1asin(nπx2a);n=2,4,6.
b) The initial wave function for a simple harmonic oscillator is
ψ ( x , 0 ) = 1 2 ψ 0 ( x ) i 3 2 ψ 2 ( x ) ψ ( x , 0 ) = 1 2 ψ 0 ( x ) i 3 2 ψ 2 ( x ) psi(x,0)=(1)/(2)psi_(0)(x)-i(sqrt3)/(2)psi_(2)(x)\psi(x, 0)=\frac{1}{2} \psi_0(x)-i \frac{\sqrt{3}}{2} \psi_2(x)ψ(x,0)=12ψ0(x)i32ψ2(x)
where ψ 0 ψ 0 psi_(0)\psi_0ψ0 and ψ 2 ψ 2 psi_(2)\psi_2ψ2 are the normalized eigenfunctions of the simple harmonic oscillator. Determine ψ ( x , t ) ψ ( x , t ) psi(x,t)\psi(x, t)ψ(x,t) and the expectation value of H ^ H ^ hat(H)\hat{H}H^ in the state ψ ( x , 0 ) ψ ( x , 0 ) psi(x,0)\psi(x, 0)ψ(x,0).
c) Determine r r (:r:)\langle r\rangler for an electron in the ψ 210 ψ 210 psi_(210)\psi_{210}ψ210 state of the hydrogen atom and show that the most probable value of r r rrr for this state is 4 a 0 4 a 0 4a_(0)4 a_04a0.
PART B
3. a) Consider the two following state vectors in a vector space spanned by the orthonormal eigenkets | ϕ 1 , | ϕ 2 , | ϕ 3 ϕ 1 , ϕ 2 , ϕ 3 |phi_(1):),|phi_(2):),|phi_(3):)\left|\phi_1\right\rangle,\left|\phi_2\right\rangle,\left|\phi_3\right\rangle|ϕ1,|ϕ2,|ϕ3 :
| ψ 1 = 2 i | ϕ 1 + | ϕ 2 i | ϕ 3 | ψ 2 = i | ϕ 1 2 | ϕ 2 ψ 1 = 2 i ϕ 1 + ϕ 2 i ϕ 3 ψ 2 = i ϕ 1 2 ϕ 2 {:[|psi_(1):)=2i|phi_(1):)+|phi_(2):)-i|phi_(3):)],[|psi_(2):)=i|phi_(1):)-2|phi_(2):)]:}\begin{aligned} & \left|\psi_1\right\rangle=2 i\left|\phi_1\right\rangle+\left|\phi_2\right\rangle-i\left|\phi_3\right\rangle \\ & \left|\psi_2\right\rangle=i\left|\phi_1\right\rangle-2\left|\phi_2\right\rangle \end{aligned}|ψ1=2i|ϕ1+|ϕ2i|ϕ3|ψ2=i|ϕ12|ϕ2
Determine
i) ψ 1 ψ 1 ; ψ 1 ψ 2 ψ 1 ψ 1 ; ψ 1 ψ 2 (:psi_(1)∣psi_(1):);(:psi_(1)∣psi_(2):)\left\langle\psi_1 \mid \psi_1\right\rangle ;\left\langle\psi_1 \mid \psi_2\right\rangleψ1ψ1;ψ1ψ2; and ψ 2 ψ 2 ψ 2 ψ 2 (:psi_(2)∣psi_(2):)\left\langle\psi_2 \mid \psi_2\right\rangleψ2ψ2
ii) The matrix elements of the operator | ψ 1 ψ 2 | ψ 1 ψ 2 |psi_(1):)(:psi_(2)|\left|\psi_1\right\rangle\left\langle\psi_2\right||ψ1ψ2|
b) Determine d x ^ d t d x ^ d t (d( hat(x)))/(dt)\frac{d \hat{x}}{d t}dx^dt and d p ^ d t d p ^ d t (d( hat(p)))/(dt)\frac{d \hat{p}}{d t}dp^dt for a particle of mass m m mmm in a gravitational field, with the Hamiltonian: H = p 2 2 m + m g x H = p 2 2 m + m g x H=(p^(2))/(2m)+mgxH=\frac{p^2}{2 m}+m g xH=p22m+mgx. Solve the equations of motion to get x ^ ( t ) x ^ ( t ) hat(x)(t)\hat{x}(t)x^(t) and p ^ ( t ) p ^ ( t ) hat(p)(t)\hat{p}(t)p^(t) in terms of x ^ ( t = 0 ) = x ^ ( 0 ) x ^ ( t = 0 ) = x ^ ( 0 ) hat(x)(t=0)= hat(x)(0)\hat{x}(t=0)=\hat{x}(0)x^(t=0)=x^(0) and p ^ ( t = 0 ) = p ^ ( 0 ) p ^ ( t = 0 ) = p ^ ( 0 ) hat(p)(t=0)= hat(p)(0)\hat{p}(t=0)=\hat{p}(0)p^(t=0)=p^(0) and calculate [ x ^ ( t 1 ) , p ^ ( t 1 ) ] x ^ t 1 , p ^ t 1 [( hat(x))(t_(1)),( hat(p))(t_(1))]\left[\hat{x}\left(t_1\right), \hat{p}\left(t_1\right)\right][x^(t1),p^(t1)].
4. a) i) Show that a ^ | n = n + 1 | n + 1 a ^ | n = n + 1 | n + 1 hat(a)^(†)|n:)=sqrt(n+1)|n+1:)\hat{a}^{\dagger}|n\rangle=\sqrt{n+1}|n+1\ranglea^|n=n+1|n+1.
ii) Using the equation ψ n + 1 ( x ) = 1 2 ( n + 1 ) [ ξ d d ξ ] ψ n ( x ) ψ n + 1 ( x ) = 1 2 ( n + 1 ) ξ d d ξ ψ n ( x ) psi_(n+1)(x)=(1)/(sqrt(2(n+1)))[xi-(d)/(d xi)]psi _(n)(x)\psi_{n+1}(x)=\frac{1}{\sqrt{2(n+1)}}\left[\xi-\frac{d}{d \xi}\right] \psi_n(x)ψn+1(x)=12(n+1)[ξddξ]ψn(x) determine ψ 3 ( x ) ψ 3 ( x ) psi_(3)(x)\psi_3(x)ψ3(x).
b) i) Write down the angular momentum states | j , m j j , m j |j,m_(j):)\left|j, m_j\right\rangle|j,mj for j = 5 2 j = 5 2 j=(5)/(2)j=\frac{5}{2}j=52 and the eigenvalue of J ^ 2 J ^ 2 hat(J)^(2)\hat{J}^2J^2 and J ^ z J ^ z hat(J)_(z)\hat{J}_zJ^z for each of these states.
ii) Obtain the angular momentum matrix J y J y J_(y)J_yJy for j = 1 j = 1 j=1j=1j=1.
c) Show that for the electron
s ^ x = 2 [ | | + | | ] s ^ x = 2 [ | | + | | ] hat(s)_(x)=(ℏ)/(2)[|uarr:)(:darr|+|darr:)(:uarr|]\hat{s}_x=\frac{\hbar}{2}[|\uparrow\rangle\langle\downarrow|+| \downarrow\rangle\langle\uparrow|]s^x=2[||+||]
and
S ^ y = i 2 [ | | + | | ] S ^ y = i 2 [ | | + | | ] hat(S)_(y)=(iℏ)/(2)[-|uarr:)(:darr|+|darr:)(:uarr|]\hat{S}_y=\frac{i \hbar}{2}[-|\uparrow\rangle\langle\downarrow|+| \downarrow\rangle\langle\uparrow|]S^y=i2[||+||]
\(b=c\:cos\:A+a\:cos\:C\)

MPH-004 Sample Solution 2024

mph-004-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

mph-004-solved-assignment-2024-ss-8e24e610-06c9-4b43-84f6-a5bf6ef5ab5c

PART A
  1. a) Calculate the average de Broglie wavelength of a nitrogen molecule at room temperature, given that the mass of nitrogen molecule is 4.65 × 10 26 k g 4.65 × 10 26 k g 4.65 xx10^(-26)kg4.65 \times 10^{-26} \mathrm{~kg}4.65×1026 kg.
Answer:
To calculate the average de Broglie wavelength of a nitrogen molecule at room temperature, we can use the formula for the de Broglie wavelength:
λ = h p λ = h p lambda=(h)/(p)\lambda = \frac{h}{p}λ=hp
where h h hhh is the Planck constant ( 6.626 × 10 34 m 2 kg s 1 6.626 × 10 34 m 2 kg s 1 6.626 xx10^(-34)” m”^(2)”kg s”^(-1)6.626 \times 10^{-34} \text{ m}^2 \text{kg s}^{-1}6.626×1034 m2kg s1) and p p ppp is the momentum of the particle.
At room temperature, the average kinetic energy of the nitrogen molecule is given by:
E k = 3 2 k B T E k = 3 2 k B T (:E_(k):)=(3)/(2)k_(B)T\langle E_k \rangle = \frac{3}{2} k_B TEk=32kBT
where k B k B k_(B)k_BkB is the Boltzmann constant ( 1.38 × 10 23 J K 1 1.38 × 10 23 J K 1 1.38 xx10^(-23)” J K”^(-1)1.38 \times 10^{-23} \text{ J K}^{-1}1.38×1023 J K1) and T T TTT is the temperature (room temperature is approximately 298 K).
The average kinetic energy can also be expressed in terms of the average momentum ( p p (:p:)\langle p \ranglep) of the nitrogen molecule:
E k = p 2 2 m E k = p 2 2 m (:E_(k):)=((:p^(2):))/(2m)\langle E_k \rangle = \frac{\langle p^2 \rangle}{2m}Ek=p22m
where m m mmm is the mass of the nitrogen molecule.
Equating the two expressions for the average kinetic energy and solving for the average momentum, we get:
p = 3 m k B T p = 3 m k B T (:p:)=sqrt(3mk_(B)T)\langle p \rangle = \sqrt{3mk_B T}p=3mkBT
Substituting this into the formula for the de Broglie wavelength, we get:
λ = h 3 m k B T λ = h 3 m k B T (:lambda:)=(h)/(sqrt(3mk_(B)T))\langle \lambda \rangle = \frac{h}{\sqrt{3mk_B T}}λ=h3mkBT
Now, we can substitute the values for h h hhh, m m mmm, k B k B k_(B)k_BkB, and T T TTT to calculate the average de Broglie wavelength of a nitrogen molecule at room temperature:
λ = 6.626 × 10 34 m 2 kg s 1 3 × 4.65 × 10 26 kg × 1.38 × 10 23 J K 1 × 298 K λ = 6.626 × 10 34 m 2 kg s 1 3 × 4.65 × 10 26 kg × 1.38 × 10 23 J K 1 × 298 K (:lambda:)=(6.626 xx10^(-34)” m”^(2)”kg s”^(-1))/(sqrt(3xx4.65 xx10^(-26)” kg”xx1.38 xx10^(-23)” J K”^(-1)xx298″ K”))\langle \lambda \rangle = \frac{6.626 \times 10^{-34} \text{ m}^2 \text{kg s}^{-1}}{\sqrt{3 \times 4.65 \times 10^{-26} \text{ kg} \times 1.38 \times 10^{-23} \text{ J K}^{-1} \times 298 \text{ K}}}λ=6.626×1034 m2kg s13×4.65×1026 kg×1.38×1023 J K1×298 K
λ = 6.626 × 10 34 3 × 4.65 × 1.38 × 298 × 10 26 23 λ = 6.626 × 10 34 3 × 4.65 × 1.38 × 298 × 10 26 23 (:lambda:)=(6.626 xx10^(-34))/(sqrt(3xx4.65 xx1.38 xx298 xx10^(-26-23)))\langle \lambda \rangle = \frac{6.626 \times 10^{-34}}{\sqrt{3 \times 4.65 \times 1.38 \times 298 \times 10^{-26-23}}}λ=6.626×10343×4.65×1.38×298×102623
λ 1.97 × 10 11 m λ 1.97 × 10 11 m (:lambda:)~~1.97 xx10^(-11)” m”\langle \lambda \rangle \approx 1.97 \times 10^{-11} \text{ m}λ1.97×1011 m
So, the average de Broglie wavelength of a nitrogen molecule at room temperature is approximately 1.97 × 10 11 m 1.97 × 10 11 m 1.97 xx10^(-11)” m”1.97 \times 10^{-11} \text{ m}1.97×1011 m or 19.7 pm 19.7 pm 19.7″ pm”19.7 \text{ pm}19.7 pm.
b) A particle of mass m m mmm is constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance L 0 L 0 L_(0)L_0L0. Using the uncertainty principle, determine the zero-point energy of the particle.
Answer:
The zero-point energy of a particle is the lowest possible energy it can have, which is a consequence of the Heisenberg uncertainty principle. For a particle constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance L 0 L 0 L_(0)L_0L0, the uncertainty in position ( Δ x Δ x Delta x\Delta xΔx) can be approximated as the width of the region, L 0 L 0 L_(0)L_0L0.
According to the Heisenberg uncertainty principle, the product of the uncertainties in position and momentum is equal to or greater than / 2 / 2 ℏ//2\hbar/2/2, where \hbar is the reduced Planck constant:
Δ x Δ p 2 Δ x Δ p 2 Delta x*Delta p >= (ℏ)/(2)\Delta x \cdot \Delta p \geq \frac{\hbar}{2}ΔxΔp2
Solving for the uncertainty in momentum ( Δ p Δ p Delta p\Delta pΔp):
Δ p 2 Δ x = 2 L 0 Δ p 2 Δ x = 2 L 0 Delta p >= (ℏ)/(2Delta x)=(ℏ)/(2L_(0))\Delta p \geq \frac{\hbar}{2\Delta x} = \frac{\hbar}{2L_0}Δp2Δx=2L0
The kinetic energy ( K K KKK) of the particle is related to its momentum ( p p ppp) by:
K = p 2 2 m K = p 2 2 m K=(p^(2))/(2m)K = \frac{p^2}{2m}K=p22m
Using the uncertainty in momentum as an estimate for the momentum of the particle, the zero-point energy ( E 0 E 0 E_(0)E_0E0) can be approximated as the minimum kinetic energy:
E 0 = ( Δ p ) 2 2 m ( 2 L 0 ) 2 2 m = 2 8 m L 0 2 E 0 = ( Δ p ) 2 2 m 2 L 0 2 2 m = 2 8 m L 0 2 E_(0)=((Delta p)^(2))/(2m) >= (((ℏ)/(2L_(0)))^(2))/(2m)=(ℏ^(2))/(8mL_(0)^(2))E_0 = \frac{(\Delta p)^2}{2m} \geq \frac{\left(\frac{\hbar}{2L_0}\right)^2}{2m} = \frac{\hbar^2}{8mL_0^2}E0=(Δp)22m(2L0)22m=28mL02
Therefore, the zero-point energy of the particle constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance L 0 L 0 L_(0)L_0L0 is given by:
E 0 2 8 m L 0 2 E 0 2 8 m L 0 2 E_(0) >= (ℏ^(2))/(8mL_(0)^(2))E_0 \geq \frac{\hbar^2}{8mL_0^2}E028mL02
This expression provides an estimate of the minimum energy that the particle can have due to the constraints imposed by the uncertainty principle and the spatial confinement between the potential barriers.

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