Uncertainty Principle UPSC Physics Optional Questions

1. Find the energy, momentum and wavelength of photon emitted by a hydrogen atom making a direct transition from an excited state with n=10 to the ground state. Also find the recoil speed of the hydrogen atom in this process.

2. Write the commutation relations for the position variable $x$ and the momentum component $p_x$,$p_y$ and $p_z$. Explain the physical significance of these relations. [Mains 2005]

3. What is the magnitude of linear momentum of a photon in a beam of He-Neon Laser $\lambda = 634 nm$. Express in terms of eV.

From De Broglie's wavelength we prove that $ \\ p = \frac{h}{\lambda} = \frac{hc}{\lambda c} \\$ $ \implies p = \frac{1240 eV nm}{(634nm)C} = 1.96 \frac{eV}{C} $

4. Use the uncertainty principle to estimate $\\$ (a) the ground state radius of hydrogen atom $\\$ (b) the ground state energy of the hydrogen atom

5. Estimate the size of hydrogen atom and the ground state energy from uncertainty principle.

Energy of H-atom is given by, $$ \tag{1} E = \frac{-Re^2}{r} + \frac{p^2}{2m_e} $$ From HUP, $ \Delta x \Delta p \ge \hbar$, $$ \tag{2} \Delta p_x \approx \frac{\hbar}{r} $$ Put (2) in (1), $$ \tag{3} E = \frac{-ke^2}{r} + \frac{\hbar^2}{2mr^2} $$ For minimum energy, $$ \frac{dE}{dr} =0 \space \& \frac{d^2E}{dr^2} < 0$$ for $ r=r_0, \implies \frac{dE}{dr} = \frac{2ke^2}{r^2} - \frac{\hbar^2}{2mr^3} = 0 \\$ $\implies r_0 = 0.52 Å$. $\\$ Note: you can put $r_0$ in $\frac{d^2E}{dr^2}$ and find E for $r=r_0$ $\\$ $\implies E = -13.6eV$

6. A typical atomic radius is about $5 \times 10^{-15} m$ and the energy of a beta particle($\beta$) emitted from the nucleus is at most $1MeV$. Prove on the basis of the uncertainty principle that electrons are not present inside a nucleus. [PYQ]

As radius of atom is about $5 \times 10^{-15}m$, so the electron to be confined inside of nucleus of uncertainty in its position should not be more than this. Hence $\Delta x = 5 \times 10^{-15}m$. Thus corresponding uncertainty in momentum is $$ \Delta p \Delta x \approx \frac{\hbar}{2} \implies \Delta p \approx \frac{\hbar}{2 \Delta x} $$ $$ \implies \frac{1.054 \times 10^{-34} Js}{10^{-14}m} = 1.054 \times 10^{-20} Kg m/s $$ Now momentum of $e^-$ inside of nucleus must be at least $p = \Delta p = 1.054 \times 10^{-20} kgm/s$. Electron with such momentum would have kinetic energy equal to $E=pc \space$(neglecting rest mass energy). Thus, $$ E =1.054 \times 10^{-20} Kgms^{-1} \times 3 \times 10^{8} ms^{-1} = 3.1514 \times 10^{-12} J = 19.6 \space MeV$$ This energy is far greater than emitted $\beta$ particle whose maxim energy was 1 MeV. Thus electron can't be part of nucleus.

7. Calculate $(\Delta x)^2$, where $\Delta x = x - \braket{x}$

We take $x$ as Hermitian operator and $\psi$ as its normalised state function. $$ (\Delta x)^2 = (x - \braket{x})^2 \\ \implies (\Delta x)^2 = x^2 + \braket{x}^2 - 2 \cdot x \cdot \braket{x} \\ \implies \braket{\psi | (\Delta x)^2 | \psi} = \braket{\psi | x^2 | \psi} + \braket{\psi | \braket{x}^2 | \psi} - 2 \braket{\psi | \cdot x \cdot \braket{x} | \psi} \\ \implies (\Delta x)^2 = \braket{x^2} + \braket{x}^2 - 2 \braket{x}^2 \\ \implies (\Delta x)^2 = \braket{x^2} - \braket{x}^2 $$

8. A beam of $4.0$ keV electrons from a source is incident on a target 50.0 cm away. Find the radius of the electron beam spot due to Heisenberg's uncertainty principle

9. Using uncertainty principle, estimate ground state energy of a simple harmonic oscillator.

10. In a series of experiments on the determination of mass of a certain elementary particle, the results showed variation of $ \pm 20 m_e$. Estimate the lifetime of the particle. ($m_e$ is the mass of electron) [PYQ]

From, HUP, we know that, $$ \boxed { \Delta E \Delta t \ge \frac{\hbar}{2} }$$ $\Delta E = \Delta m c^2$ $\\$ $\Delta E = (40m_e)c^2$ $\\$ (since variation in mass is $\pm20m_e$)

11. If $\hat{x}$ and $\hat{p}$ are the position and momentum operators, prove the commutation relation$ \left[\hat{p}^2,\hat{x}\right]=-2i \hbar p$ [Mains 2014]