# IGNOU MPH-004 Solved Assignment 2024 | MSCPH | IGNOU

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

365.00

Access via our Android App Only

Details For MPH-004 Solved Assignment

## 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}\text{}\mathrm{k}\mathrm{g}$$4.65×{10}^{-26}\text{}\mathrm{k}\mathrm{g}$4.65 xx10^(-26)kg4.65 \times 10^{-26} \mathrm{~kg}$4.65×{10}^{-26}\text{}\mathrm{k}\mathrm{g}$.
b) A particle of mass $m$$m$mm$m$ is constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance ${L}_{0}$${L}_{0}$L_(0)L_0${L}_{0}$. Using the uncertainty principle, determine the zero-point energy of the particle.
c) The wavefunction for a particle is given by:
$\psi \left(x\right)=\left\{\begin{array}{cc}N\left({L}^{2}-{x}^{2}\right)& -L\le x\le L\\ 0& \text{elsewhere}\end{array}$$\psi \left(x\right)=\left\{\begin{array}{cc}N\left({L}^{2}-{x}^{2}\right)& -L\le x\le L\\ 0& \text{elsewhere}\end{array}\right\$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.$\psi \left(x\right)=\left\{\begin{array}{cc}N\left({L}^{2}-{x}^{2}\right)& -L\le x\le L\\ 0& \text{elsewhere}\end{array}$
Calculate the normalization constant $N$$N$NN$N$.
d) Show that $\left[{\stackrel{^}{L}}_{x}{\stackrel{^}{L}}_{y},{\stackrel{^}{L}}_{z}\right]=i\hslash \left({\stackrel{^}{L}}_{x}{}^{2}-{\stackrel{^}{L}}_{y}{}^{2}\right)$$\left[{\stackrel{^}{L}}_{x}{\stackrel{^}{L}}_{y},{\stackrel{^}{L}}_{z}\right]=i\hslash \left({\stackrel{^}{L}}_{x}{}^{2}-{\stackrel{^}{L}}_{y}{}^{2}\right)$[ hat(L)_(x) hat(L)_(y), hat(L)_(z)]=iℏ( hat(L)_(x)^(2)- hat(L)_(y)^(2))\left[\hat{L}_x \hat{L}_y, \hat{L}_z\right]=i \hbar\left(\hat{L}_x{ }^2-\hat{L}_y{ }^2\right)$\left[{\stackrel{^}{L}}_{x}{\stackrel{^}{L}}_{y},{\stackrel{^}{L}}_{z}\right]=i\hslash \left({\stackrel{^}{L}}_{x}{}^{2}-{\stackrel{^}{L}}_{y}{}^{2}\right)$
2. a) Calculate the expectation values $⟨\stackrel{^}{x}⟩$$⟨\stackrel{^}{x}⟩$(: hat(x):)\langle\hat{x}\rangle$⟨\stackrel{^}{x}⟩$ and $⟨{\stackrel{^}{x}}^{2}⟩$$⟨{\stackrel{^}{x}}^{2}⟩$(: hat(x)^(2):)\left\langle\hat{x}^2\right\rangle$⟨{\stackrel{^}{x}}^{2}⟩$ for the following odd parity state of a symmetric infinite potential well
$\psi \left(x\right)=\frac{1}{\sqrt{a}}\mathrm{sin}\left(\frac{n\pi x}{2a}\right);n=2,4,6\dots .$$\psi \left(x\right)=\frac{1}{\sqrt{a}}\mathrm{sin}\left(\frac{n\pi x}{2a}\right);n=2,4,6\dots .$psi(x)=(1)/(sqrta)sin((n pi x)/(2a));n=2,4,6dots.\psi(x)=\frac{1}{\sqrt{a}} \sin \left(\frac{n \pi x}{2 a}\right) ; n=2,4,6 \ldots .$\psi \left(x\right)=\frac{1}{\sqrt{a}}\mathrm{sin}\left(\frac{n\pi x}{2a}\right);n=2,4,6\dots .$
b) The initial wave function for a simple harmonic oscillator is
$\psi \left(x,0\right)=\frac{1}{2}{\psi }_{0}\left(x\right)-i\frac{\sqrt{3}}{2}{\psi }_{2}\left(x\right)$$\psi \left(x,0\right)=\frac{1}{2}{\psi }_{0}\left(x\right)-i\frac{\sqrt{3}}{2}{\psi }_{2}\left(x\right)$psi(x,0)=(1)/(2)psi_(0)(x)-i(sqrt3)/(2)psi_(2)(x)\psi(x, 0)=\frac{1}{2} \psi_0(x)-i \frac{\sqrt{3}}{2} \psi_2(x)$\psi \left(x,0\right)=\frac{1}{2}{\psi }_{0}\left(x\right)-i\frac{\sqrt{3}}{2}{\psi }_{2}\left(x\right)$
where ${\psi }_{0}$${\psi }_{0}$psi_(0)\psi_0${\psi }_{0}$ and ${\psi }_{2}$${\psi }_{2}$psi_(2)\psi_2${\psi }_{2}$ are the normalized eigenfunctions of the simple harmonic oscillator. Determine $\psi \left(x,t\right)$$\psi \left(x,t\right)$psi(x,t)\psi(x, t)$\psi \left(x,t\right)$ and the expectation value of $\stackrel{^}{H}$$\stackrel{^}{H}$hat(H)\hat{H}$\stackrel{^}{H}$ in the state $\psi \left(x,0\right)$$\psi \left(x,0\right)$psi(x,0)\psi(x, 0)$\psi \left(x,0\right)$.
c) Determine $⟨r⟩$$⟨r⟩$(:r:)\langle r\rangle$⟨r⟩$ for an electron in the ${\psi }_{210}$${\psi }_{210}$psi_(210)\psi_{210}${\psi }_{210}$ state of the hydrogen atom and show that the most probable value of $r$$r$rr$r$ for this state is $4{a}_{0}$$4{a}_{0}$4a_(0)4 a_0$4{a}_{0}$.
PART B
3. a) Consider the two following state vectors in a vector space spanned by the orthonormal eigenkets $|{\varphi }_{1}⟩,|{\varphi }_{2}⟩,|{\varphi }_{3}⟩$$\left|{\varphi }_{1}\right⟩,\left|{\varphi }_{2}\right⟩,\left|{\varphi }_{3}\right⟩$|phi_(1):),|phi_(2):),|phi_(3):)\left|\phi_1\right\rangle,\left|\phi_2\right\rangle,\left|\phi_3\right\rangle$|{\varphi }_{1}⟩,|{\varphi }_{2}⟩,|{\varphi }_{3}⟩$ :
$\begin{array}{rl}& |{\psi }_{1}⟩=2i|{\varphi }_{1}⟩+|{\varphi }_{2}⟩-i|{\varphi }_{3}⟩\\ & |{\psi }_{2}⟩=i|{\varphi }_{1}⟩-2|{\varphi }_{2}⟩\end{array}$$\begin{array}{r}\left|{\psi }_{1}\right⟩=2i\left|{\varphi }_{1}\right⟩+\left|{\varphi }_{2}\right⟩-i\left|{\varphi }_{3}\right⟩\\ \left|{\psi }_{2}\right⟩=i\left|{\varphi }_{1}\right⟩-2\left|{\varphi }_{2}\right⟩\end{array}${:[|psi_(1):)=2i|phi_(1):)+|phi_(2):)-i|phi_(3):)],[|psi_(2):)=i|phi_(1):)-2|phi_(2):)]:}\begin{aligned} & \left|\psi_1\right\rangle=2 i\left|\phi_1\right\rangle+\left|\phi_2\right\rangle-i\left|\phi_3\right\rangle \\ & \left|\psi_2\right\rangle=i\left|\phi_1\right\rangle-2\left|\phi_2\right\rangle \end{aligned}$\begin{array}{rl}& |{\psi }_{1}⟩=2i|{\varphi }_{1}⟩+|{\varphi }_{2}⟩-i|{\varphi }_{3}⟩\\ & |{\psi }_{2}⟩=i|{\varphi }_{1}⟩-2|{\varphi }_{2}⟩\end{array}$
Determine
i) $⟨{\psi }_{1}\mid {\psi }_{1}⟩;⟨{\psi }_{1}\mid {\psi }_{2}⟩$$⟨{\psi }_{1}\mid {\psi }_{1}⟩;⟨{\psi }_{1}\mid {\psi }_{2}⟩$(:psi_(1)∣psi_(1):);(:psi_(1)∣psi_(2):)\left\langle\psi_1 \mid \psi_1\right\rangle ;\left\langle\psi_1 \mid \psi_2\right\rangle$⟨{\psi }_{1}\mid {\psi }_{1}⟩;⟨{\psi }_{1}\mid {\psi }_{2}⟩$; and $⟨{\psi }_{2}\mid {\psi }_{2}⟩$$⟨{\psi }_{2}\mid {\psi }_{2}⟩$(:psi_(2)∣psi_(2):)\left\langle\psi_2 \mid \psi_2\right\rangle$⟨{\psi }_{2}\mid {\psi }_{2}⟩$
ii) The matrix elements of the operator $|{\psi }_{1}⟩⟨{\psi }_{2}|$$\left|{\psi }_{1}\right⟩⟨{\psi }_{2}|$|psi_(1):)(:psi_(2)|\left|\psi_1\right\rangle\left\langle\psi_2\right|$|{\psi }_{1}⟩⟨{\psi }_{2}|$
b) Determine $\frac{d\stackrel{^}{x}}{dt}$$\frac{d\stackrel{^}{x}}{dt}$(d( hat(x)))/(dt)\frac{d \hat{x}}{d t}$\frac{d\stackrel{^}{x}}{dt}$ and $\frac{d\stackrel{^}{p}}{dt}$$\frac{d\stackrel{^}{p}}{dt}$(d( hat(p)))/(dt)\frac{d \hat{p}}{d t}$\frac{d\stackrel{^}{p}}{dt}$ for a particle of mass $m$$m$mm$m$ in a gravitational field, with the Hamiltonian: $H=\frac{{p}^{2}}{2m}+mgx$$H=\frac{{p}^{2}}{2m}+mgx$H=(p^(2))/(2m)+mgxH=\frac{p^2}{2 m}+m g x$H=\frac{{p}^{2}}{2m}+mgx$. Solve the equations of motion to get $\stackrel{^}{x}\left(t\right)$$\stackrel{^}{x}\left(t\right)$hat(x)(t)\hat{x}(t)$\stackrel{^}{x}\left(t\right)$ and $\stackrel{^}{p}\left(t\right)$$\stackrel{^}{p}\left(t\right)$hat(p)(t)\hat{p}(t)$\stackrel{^}{p}\left(t\right)$ in terms of $\stackrel{^}{x}\left(t=0\right)=\stackrel{^}{x}\left(0\right)$$\stackrel{^}{x}\left(t=0\right)=\stackrel{^}{x}\left(0\right)$hat(x)(t=0)= hat(x)(0)\hat{x}(t=0)=\hat{x}(0)$\stackrel{^}{x}\left(t=0\right)=\stackrel{^}{x}\left(0\right)$ and $\stackrel{^}{p}\left(t=0\right)=\stackrel{^}{p}\left(0\right)$$\stackrel{^}{p}\left(t=0\right)=\stackrel{^}{p}\left(0\right)$hat(p)(t=0)= hat(p)(0)\hat{p}(t=0)=\hat{p}(0)$\stackrel{^}{p}\left(t=0\right)=\stackrel{^}{p}\left(0\right)$ and calculate $\left[\stackrel{^}{x}\left({t}_{1}\right),\stackrel{^}{p}\left({t}_{1}\right)\right]$$\left[\stackrel{^}{x}\left({t}_{1}\right),\stackrel{^}{p}\left({t}_{1}\right)\right]$[( hat(x))(t_(1)),( hat(p))(t_(1))]\left[\hat{x}\left(t_1\right), \hat{p}\left(t_1\right)\right]$\left[\stackrel{^}{x}\left({t}_{1}\right),\stackrel{^}{p}\left({t}_{1}\right)\right]$.
4. a) i) Show that ${\stackrel{^}{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$${\stackrel{^}{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$hat(a)^(†)|n:)=sqrt(n+1)|n+1:)\hat{a}^{\dagger}|n\rangle=\sqrt{n+1}|n+1\rangle${\stackrel{^}{a}}^{†}|n⟩=\sqrt{n+1}|n+1⟩$.
ii) Using the equation ${\psi }_{n+1}\left(x\right)=\frac{1}{\sqrt{2\left(n+1\right)}}\left[\xi -\frac{d}{d\xi }\right]{\psi }_{n}\left(x\right)$${\psi }_{n+1}\left(x\right)=\frac{1}{\sqrt{2\left(n+1\right)}}\left[\xi -\frac{d}{d\xi }\right]{\psi }_{n}\left(x\right)$psi_(n+1)(x)=(1)/(sqrt(2(n+1)))[xi-(d)/(d xi)]psi _(n)(x)\psi_{n+1}(x)=\frac{1}{\sqrt{2(n+1)}}\left[\xi-\frac{d}{d \xi}\right] \psi_n(x)${\psi }_{n+1}\left(x\right)=\frac{1}{\sqrt{2\left(n+1\right)}}\left[\xi -\frac{d}{d\xi }\right]{\psi }_{n}\left(x\right)$ determine ${\psi }_{3}\left(x\right)$${\psi }_{3}\left(x\right)$psi_(3)(x)\psi_3(x)${\psi }_{3}\left(x\right)$.
b) i) Write down the angular momentum states $|j,{m}_{j}⟩$$\left|j,{m}_{j}\right⟩$|j,m_(j):)\left|j, m_j\right\rangle$|j,{m}_{j}⟩$ for $j=\frac{5}{2}$$j=\frac{5}{2}$j=(5)/(2)j=\frac{5}{2}$j=\frac{5}{2}$ and the eigenvalue of ${\stackrel{^}{J}}^{2}$${\stackrel{^}{J}}^{2}$hat(J)^(2)\hat{J}^2${\stackrel{^}{J}}^{2}$ and ${\stackrel{^}{J}}_{z}$${\stackrel{^}{J}}_{z}$hat(J)_(z)\hat{J}_z${\stackrel{^}{J}}_{z}$ for each of these states.
ii) Obtain the angular momentum matrix ${J}_{y}$${J}_{y}$J_(y)J_y${J}_{y}$ for $j=1$$j=1$j=1j=1$j=1$.
c) Show that for the electron
${\stackrel{^}{s}}_{x}=\frac{\hslash }{2}\left[|↑⟩⟨↓|+|↓⟩⟨↑|\right]$${\stackrel{^}{s}}_{x}=\frac{\hslash }{2}\left[|↑⟩⟨↓|+|↓⟩⟨↑|\right]$hat(s)_(x)=(ℏ)/(2)[|uarr:)(:darr|+|darr:)(:uarr|]\hat{s}_x=\frac{\hbar}{2}[|\uparrow\rangle\langle\downarrow|+| \downarrow\rangle\langle\uparrow|]${\stackrel{^}{s}}_{x}=\frac{\hslash }{2}\left[|↑⟩⟨↓|+|↓⟩⟨↑|\right]$
and
${\stackrel{^}{S}}_{y}=\frac{i\hslash }{2}\left[-|↑⟩⟨↓|+|↓⟩⟨↑|\right]$${\stackrel{^}{S}}_{y}=\frac{i\hslash }{2}\left[-|↑⟩⟨↓|+|↓⟩⟨↑|\right]$hat(S)_(y)=(iℏ)/(2)[-|uarr:)(:darr|+|darr:)(:uarr|]\hat{S}_y=\frac{i \hbar}{2}[-|\uparrow\rangle\langle\downarrow|+| \downarrow\rangle\langle\uparrow|]${\stackrel{^}{S}}_{y}=\frac{i\hslash }{2}\left[-|↑⟩⟨↓|+|↓⟩⟨↑|\right]$
$$b=c\:cos\:A+a\:cos\:C$$

## 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}\text{}\mathrm{k}\mathrm{g}$$4.65×{10}^{-26}\text{}\mathrm{k}\mathrm{g}$4.65 xx10^(-26)kg4.65 \times 10^{-26} \mathrm{~kg}$4.65×{10}^{-26}\text{}\mathrm{k}\mathrm{g}$.
To calculate the average de Broglie wavelength of a nitrogen molecule at room temperature, we can use the formula for the de Broglie wavelength:
$\lambda =\frac{h}{p}$$\lambda =\frac{h}{p}$lambda=(h)/(p)\lambda = \frac{h}{p}$\lambda =\frac{h}{p}$
where $h$$h$hh$h$ is the Planck constant ($6.626×{10}^{-34}{\text{m}}^{2}{\text{kg s}}^{-1}$$6.626×{10}^{-34}{\text{m}}^{2}{\text{kg s}}^{-1}$6.626 xx10^(-34)” m”^(2)”kg s”^(-1)6.626 \times 10^{-34} \text{ m}^2 \text{kg s}^{-1}$6.626×{10}^{-34}{\text{m}}^{2}{\text{kg s}}^{-1}$) and $p$$p$pp$p$ is the momentum of the particle.
At room temperature, the average kinetic energy of the nitrogen molecule is given by:
$⟨{E}_{k}⟩=\frac{3}{2}{k}_{B}T$$⟨{E}_{k}⟩=\frac{3}{2}{k}_{B}T$(:E_(k):)=(3)/(2)k_(B)T\langle E_k \rangle = \frac{3}{2} k_B T$⟨{E}_{k}⟩=\frac{3}{2}{k}_{B}T$
where ${k}_{B}$${k}_{B}$k_(B)k_B${k}_{B}$ is the Boltzmann constant ($1.38×{10}^{-23}{\text{J K}}^{-1}$$1.38×{10}^{-23}{\text{J K}}^{-1}$1.38 xx10^(-23)” J K”^(-1)1.38 \times 10^{-23} \text{ J K}^{-1}$1.38×{10}^{-23}{\text{J K}}^{-1}$) and $T$$T$TT$T$ is the temperature (room temperature is approximately 298 K).
The average kinetic energy can also be expressed in terms of the average momentum ($⟨p⟩$$⟨p⟩$(:p:)\langle p \rangle$⟨p⟩$) of the nitrogen molecule:
$⟨{E}_{k}⟩=\frac{⟨{p}^{2}⟩}{2m}$$⟨{E}_{k}⟩=\frac{⟨{p}^{2}⟩}{2m}$(:E_(k):)=((:p^(2):))/(2m)\langle E_k \rangle = \frac{\langle p^2 \rangle}{2m}$⟨{E}_{k}⟩=\frac{⟨{p}^{2}⟩}{2m}$
where $m$$m$mm$m$ 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⟩=\sqrt{3m{k}_{B}T}$$⟨p⟩=\sqrt{3m{k}_{B}T}$(:p:)=sqrt(3mk_(B)T)\langle p \rangle = \sqrt{3mk_B T}$⟨p⟩=\sqrt{3m{k}_{B}T}$
Substituting this into the formula for the de Broglie wavelength, we get:
$⟨\lambda ⟩=\frac{h}{\sqrt{3m{k}_{B}T}}$$⟨\lambda ⟩=\frac{h}{\sqrt{3m{k}_{B}T}}$(:lambda:)=(h)/(sqrt(3mk_(B)T))\langle \lambda \rangle = \frac{h}{\sqrt{3mk_B T}}$⟨\lambda ⟩=\frac{h}{\sqrt{3m{k}_{B}T}}$
Now, we can substitute the values for $h$$h$hh$h$, $m$$m$mm$m$, ${k}_{B}$${k}_{B}$k_(B)k_B${k}_{B}$, and $T$$T$TT$T$ to calculate the average de Broglie wavelength of a nitrogen molecule at room temperature:
$⟨\lambda ⟩=\frac{6.626×{10}^{-34}{\text{m}}^{2}{\text{kg s}}^{-1}}{\sqrt{3×4.65×{10}^{-26}\text{kg}×1.38×{10}^{-23}{\text{J K}}^{-1}×298\text{K}}}$$⟨\lambda ⟩=\frac{6.626×{10}^{-34}{\text{m}}^{2}{\text{kg s}}^{-1}}{\sqrt{3×4.65×{10}^{-26}\text{kg}×1.38×{10}^{-23}{\text{J K}}^{-1}×298\text{K}}}$(:lambda:)=(6.626 xx10^(-34)” m”^(2)”kg s”^(-1))/(sqrt(3xx4.65 xx10^(-26)” kg”xx1.38 xx10^(-23)” J K”^(-1)xx298″ K”))\langle \lambda \rangle = \frac{6.626 \times 10^{-34} \text{ m}^2 \text{kg s}^{-1}}{\sqrt{3 \times 4.65 \times 10^{-26} \text{ kg} \times 1.38 \times 10^{-23} \text{ J K}^{-1} \times 298 \text{ K}}}$⟨\lambda ⟩=\frac{6.626×{10}^{-34}{\text{m}}^{2}{\text{kg s}}^{-1}}{\sqrt{3×4.65×{10}^{-26}\text{kg}×1.38×{10}^{-23}{\text{J K}}^{-1}×298\text{K}}}$
$⟨\lambda ⟩=\frac{6.626×{10}^{-34}}{\sqrt{3×4.65×1.38×298×{10}^{-26-23}}}$$⟨\lambda ⟩=\frac{6.626×{10}^{-34}}{\sqrt{3×4.65×1.38×298×{10}^{-26-23}}}$(:lambda:)=(6.626 xx10^(-34))/(sqrt(3xx4.65 xx1.38 xx298 xx10^(-26-23)))\langle \lambda \rangle = \frac{6.626 \times 10^{-34}}{\sqrt{3 \times 4.65 \times 1.38 \times 298 \times 10^{-26-23}}}$⟨\lambda ⟩=\frac{6.626×{10}^{-34}}{\sqrt{3×4.65×1.38×298×{10}^{-26-23}}}$
$⟨\lambda ⟩\approx 1.97×{10}^{-11}\text{m}$$⟨\lambda ⟩\approx 1.97×{10}^{-11}\text{m}$(:lambda:)~~1.97 xx10^(-11)” m”\langle \lambda \rangle \approx 1.97 \times 10^{-11} \text{ m}$⟨\lambda ⟩\approx 1.97×{10}^{-11}\text{m}$
So, the average de Broglie wavelength of a nitrogen molecule at room temperature is approximately $1.97×{10}^{-11}\text{m}$$1.97×{10}^{-11}\text{m}$1.97 xx10^(-11)” m”1.97 \times 10^{-11} \text{ m}$1.97×{10}^{-11}\text{m}$ or $19.7\text{pm}$$19.7\text{pm}$19.7″ pm”19.7 \text{ pm}$19.7\text{pm}$.
b) A particle of mass $m$$m$mm$m$ is constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance ${L}_{0}$${L}_{0}$L_(0)L_0${L}_{0}$. Using the uncertainty principle, determine the zero-point energy of the particle.
The zero-point energy of a particle is the lowest possible energy it can have, which is a consequence of the Heisenberg uncertainty principle. For a particle constrained to move in a one-dimensional region between two infinitely high potential barriers separated by a distance ${L}_{0}$${L}_{0}$L_(0)L_0${L}_{0}$, the uncertainty in position ($\mathrm{\Delta }x$$\mathrm{\Delta }x$Delta x\Delta x$\mathrm{\Delta }x$) can be approximated as the width of the region, ${L}_{0}$${L}_{0}$L_(0)L_0${L}_{0}$.
According to the Heisenberg uncertainty principle, the product of the uncertainties in position and momentum is equal to or greater than $\hslash /2$$\hslash /2$ℏ//2\hbar/2$\hslash /2$, where $\hslash$$\hslash$\hbar$\hslash$ is the reduced Planck constant:
$\mathrm{\Delta }x\cdot \mathrm{\Delta }p\ge \frac{\hslash }{2}$$\mathrm{\Delta }x\cdot \mathrm{\Delta }p\ge \frac{\hslash }{2}$Delta x*Delta p >= (ℏ)/(2)\Delta x \cdot \Delta p \geq \frac{\hbar}{2}$\mathrm{\Delta }x\cdot \mathrm{\Delta }p\ge \frac{\hslash }{2}$
Solving for the uncertainty in momentum ($\mathrm{\Delta }p$$\mathrm{\Delta }p$Delta p\Delta p$\mathrm{\Delta }p$):
$\mathrm{\Delta }p\ge \frac{\hslash }{2\mathrm{\Delta }x}=\frac{\hslash }{2{L}_{0}}$$\mathrm{\Delta }p\ge \frac{\hslash }{2\mathrm{\Delta }x}=\frac{\hslash }{2{L}_{0}}$Delta p >= (ℏ)/(2Delta x)=(ℏ)/(2L_(0))\Delta p \geq \frac{\hbar}{2\Delta x} = \frac{\hbar}{2L_0}$\mathrm{\Delta }p\ge \frac{\hslash }{2\mathrm{\Delta }x}=\frac{\hslash }{2{L}_{0}}$
The kinetic energy ($K$$K$KK$K$) of the particle is related to its momentum ($p$$p$pp$p$) by:
$K=\frac{{p}^{2}}{2m}$$K=\frac{{p}^{2}}{2m}$K=(p^(2))/(2m)K = \frac{p^2}{2m}$K=\frac{{p}^{2}}{2m}$
Using the uncertainty in momentum as an estimate for the momentum of the particle, the zero-point energy (${E}_{0}$${E}_{0}$E_(0)E_0${E}_{0}$) can be approximated as the minimum kinetic energy:
${E}_{0}=\frac{\left(\mathrm{\Delta }p{\right)}^{2}}{2m}\ge \frac{{\left(\frac{\hslash }{2{L}_{0}}\right)}^{2}}{2m}=\frac{{\hslash }^{2}}{8m{L}_{0}^{2}}$${E}_{0}=\frac{\left(\mathrm{\Delta }p{\right)}^{2}}{2m}\ge \frac{{\left(\frac{\hslash }{2{L}_{0}}\right)}^{2}}{2m}=\frac{{\hslash }^{2}}{8m{L}_{0}^{2}}$E_(0)=((Delta p)^(2))/(2m) >= (((ℏ)/(2L_(0)))^(2))/(2m)=(ℏ^(2))/(8mL_(0)^(2))E_0 = \frac{(\Delta p)^2}{2m} \geq \frac{\left(\frac{\hbar}{2L_0}\right)^2}{2m} = \frac{\hbar^2}{8mL_0^2}${E}_{0}=\frac{\left(\mathrm{\Delta }p{\right)}^{2}}{2m}\ge \frac{{\left(\frac{\hslash }{2{L}_{0}}\right)}^{2}}{2m}=\frac{{\hslash }^{2}}{8m{L}_{0}^{2}}$
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}$${L}_{0}$L_(0)L_0${L}_{0}$ is given by:
${E}_{0}\ge \frac{{\hslash }^{2}}{8m{L}_{0}^{2}}$${E}_{0}\ge \frac{{\hslash }^{2}}{8m{L}_{0}^{2}}$E_(0) >= (ℏ^(2))/(8mL_(0)^(2))E_0 \geq \frac{\hbar^2}{8mL_0^2}${E}_{0}\ge \frac{{\hslash }^{2}}{8m{L}_{0}^{2}}$
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.

Simply click “Install” to download and install the app, and then follow the instructions to purchase the required assignment solution. Currently, the app is only available for Android devices. We are working on making the app available for iOS in the future, but it is not currently available for iOS devices.

Yes, It is Complete Solution, a comprehensive solution to the assignments for IGNOU. Valid from January 1, 2023 to December 31, 2023.

Yes, the Complete Solution is aligned with the IGNOU requirements and has been solved accordingly.

Yes, the Complete Solution is guaranteed to be error-free.The solutions are thoroughly researched and verified by subject matter experts to ensure their accuracy.

As of now, you have access to the Complete Solution for a period of 6 months after the date of purchase, which is sufficient to complete the assignment. However, we can extend the access period upon request. You can access the solution anytime through our app.

The app provides complete solutions for all assignment questions. If you still need help, you can contact the support team for assistance at Whatsapp +91-9958288900

No, access to the educational materials is limited to one device only, where you have first logged in. Logging in on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Payments can be made through various secure online payment methods available in the app.Your payment information is protected with industry-standard security measures to ensure its confidentiality and safety. You will receive a receipt for your payment through email or within the app, depending on your preference.

The instructions for formatting your assignments are detailed in the Assignment Booklet, which includes details on paper size, margins, precision, and submission requirements. It is important to strictly follow these instructions to facilitate evaluation and avoid delays.

$$c=a\:cos\:B+b\:cos\:A$$

## Terms and Conditions

• The educational materials provided in the app are the sole property of the app owner and are protected by copyright laws.
• Reproduction, distribution, or sale of the educational materials without prior written consent from the app owner is strictly prohibited and may result in legal consequences.
• Any attempt to modify, alter, or use the educational materials for commercial purposes is strictly prohibited.
• The app owner reserves the right to revoke access to the educational materials at any time without notice for any violation of these terms and conditions.
• The app owner is not responsible for any damages or losses resulting from the use of the educational materials.
• The app owner reserves the right to modify these terms and conditions at any time without notice.
• By accessing and using the app, you agree to abide by these terms and conditions.
• Access to the educational materials is limited to one device only. Logging in to the app on multiple devices is not allowed and may result in the revocation of access to the educational materials.

Our educational materials are solely available on our website and application only. Users and students can report the dealing or selling of the copied version of our educational materials by any third party at our email address (abstract4math@gmail.com) or mobile no. (+91-9958288900).

In return, such users/students can expect free our educational materials/assignments and other benefits as a bonafide gesture which will be completely dependent upon our discretion.

Scroll to Top
Scroll to Top