Atoms Set-5

Q1. A hydrogen atom makes a transition from $n=4$ to $n=2$. Using the Rydberg constant $R=1.097\times10^{7}\ \mathrm{m^{-1}}$, the wavelength of the emitted photon is approximately:

Q2. For the transition $n=3\to1$ in a hydrogen-like ion the wavenumber is $1/\lambda = R_\infty(\mu/m_e)Z^2\left(\dfrac{1}{1^2}-\dfrac{1}{3^2}\right)$ where $\mu$ is the reduced mass. Taking $M_p=1836\,m_e$ for the proton and $M_{He}=4\times1836\,m_e$ for the helium nucleus, the ratio $\lambda_H/\lambda_{He^+}$ for this transition is closest to:

Q3. Taking reduced mass into account (proton mass $1836\,m_e$, helium nucleus $7344\,m_e$) and using Bohr quantization, the ratio of the orbital speeds $v_{He^+,\,n=2}/v_{H,\,n=1}$ is:

Q4. Assertion (A): For the transition $(n+1)\to n$ in hydrogen the emitted-photon frequency $\nu_{\rm em}=\Delta E/h$ approaches the classical orbital frequency $\nu_{\rm orb}=v/(2\pi r)$ of the electron in the $(n+1)$th orbit only in the limit $n\gg1$.

Reason (R): Bohr's frequency condition $\nu_{\rm em}=\Delta E/h$ together with classical orbital relations implies $\nu_{\rm em}\to\nu_{\rm orb}$ as $n\to\infty$ (correspondence principle).

Q5. The Hα line ($n=3\to2$) of hydrogen and deuterium are slightly shifted because the reduced mass differs. With $M_p=1836\,m_e$ and $M_d=3672\,m_e$, the minimum resolving power $R=\lambda/\Delta\lambda$ required to resolve the Hα lines of H and D is approximately:

Q6. A hydrogen atom makes a transition from $n=4$ to $n=2$. Using the Rydberg formula, the wavelength of the emitted photon (take $R_H=1.097\times10^7\ \mathrm{m^{-1}}$) is closest to:

Q7. An electron with kinetic energy $15\ \mathrm{eV}$ collides with a hydrogen atom in the ground state and excites it to $n=3$ (inelastic collision, neglect recoil). Using $1\ \mathrm{eV}=1.602\times10^{-19}\ \mathrm{J}$ and $m_e=9.11\times10^{-31}\ \mathrm{kg}$, the speed of the scattered electron is approximately:

Q8. A hydrogen atom transitions from $n=4\to n=2$ producing a photon of wavelength $\lambda_1$, and another hydrogen atom transitions from $n=3\to n=1$ producing wavelength $\lambda_2$. The ratio $\lambda_1/\lambda_2$ equals:

Q9. Consider the Lyman-$\alpha$ line ($n=2\to1$) of hydrogen. If one uses the reduced mass $\mu=\dfrac{m_e m_p}{m_e+m_p}$ instead of assuming infinite nuclear mass, by approximately how much does the wavelength change compared to the infinite-mass value (take $m_e=9.11\times10^{-31}\ \mathrm{kg}$, $m_p=1.673\times10^{-27}\ \mathrm{kg}$ and Lyman-$\alpha\approx121.6\ \mathrm{nm}$ for infinite mass)?

Q10. Using the Heisenberg uncertainty estimate $\Delta p\approx \hbar/r$ and approximating the energy of an electron in hydrogen as $E(r)\approx \dfrac{\hbar^2}{2mr^2}-\dfrac{e^2}{4\pi\epsilon_0 r}$, minimize $E(r)$ to estimate the ground-state radius. Using $\hbar=1.055\times10^{-34}\ \mathrm{J\cdot s}$, $m_e=9.11\times10^{-31}\ \mathrm{kg}$, $e=1.602\times10^{-19}\ \mathrm{C}$ and $\epsilon_0=8.854\times10^{-12}\ \mathrm{F/m}$, the estimate is closest to:

Q11. An electron in a hydrogen atom makes a transition from $n=4$ to $n=2$. Using the Rydberg constant $R=1.097\times10^{7}\ \mathrm{m^{-1}}$, the wavelength $\lambda$ of the emitted photon is approximately

Q12. In the Bohr model $r_n=\dfrac{n^2 a_0}{Z}$. The orbital radius of hydrogen in $n=3$ equals $9a_0$. For the hydrogen-like ion He$^+$ ($Z=2$), the principal quantum number $n'$ for which its orbital radius equals that hydrogen $n=3$ radius is

Q13. The reduced-mass correction causes a small isotope shift between hydrogen (H) and deuterium (D). For the Balmer-$\alpha$ line ($3\to2$) take $m_p\approx1836\,m_e$ and $\lambda_H\approx656.3\ \mathrm{nm}$. Using a first-order approximation in $m_e/m_p$, the approximate difference $\Delta\lambda=\lambda_H-\lambda_D$ is

Q14. During the $2\to1$ transition in hydrogen the emitted photon has energy $\Delta E\approx10.2\ \mathrm{eV}$. Accounting for recoil of the proton, the fraction $f$ of the transition energy that becomes kinetic energy of the proton can be estimated by $f\simeq\dfrac{\Delta E}{2Mc^2}$. Using $M\simeq m_p$ and $m_p c^2\approx9.38\times10^{8}\ \mathrm{eV}$, the closest value of $f$ is

Q15. Assertion (A): In the Bohr model the ground-state electron speed is $v=Z\alpha c$, where $\alpha$ is the fine-structure constant; therefore for $Z>137$ the formula would predict $v>c$, indicating breakdown of the non-relativistic Bohr model for such large $Z$.
Reason (R): $\displaystyle \alpha=\frac{e^2}{4\pi\varepsilon_0\hbar c}\approx\frac{1}{137}$ and hence $v=Z\alpha c$ in Bohr theory.