Dual Nature of Radiation and Matter Set-4

Q1. A metal surface is illuminated by light of frequency $1.5\times10^{15}\ \mathrm{Hz}$. The stopping potential is measured to be $2.0\ \mathrm{V}$. Using Einstein's photoelectric equation $eV_0=hf-\phi$ and $h=6.63\times10^{-34}\ \mathrm{Js}$, $1\ \mathrm{eV}=1.60\times10^{-19}\ \mathrm{J}$, the work function $\phi$ of the metal is approximately:

Q2. Electrons are accelerated through a potential difference $V$ and diffract from crystal planes with interplanar spacing $d=0.20\ \mathrm{nm}$. The first-order Bragg maximum ($n=1$) is observed at angle $\theta=30^\circ$ (with respect to the planes). Using $2d\sin\theta=n\lambda$ and the de Broglie relation $\lambda=\dfrac{h}{\sqrt{2m_e eV}}$, and $h=6.63\times10^{-34}\ \mathrm{Js}$, $m_e=9.11\times10^{-31}\ \mathrm{kg}$, $e=1.60\times10^{-19}\ \mathrm{C}$, the accelerating potential $V$ is closest to:

Q3. Monochromatic light of wavelength $\lambda=400\ \mathrm{nm}$ and intensity $I=2.0\times10^{-3}\ \mathrm{W/m^2}$ falls uniformly on a photocathode of area $1.0\ \mathrm{cm^2}$. If each photon can eject at most one electron and the quantum efficiency is $20\%$, the maximum photoelectric current produced (assuming all emitted electrons are collected) is approximately. (Use $E_{\rm ph}=hc/\lambda$, $h=6.63\times10^{-34}\ \mathrm{Js}$, $c=3.0\times10^8\ \mathrm{m/s}$, $e=1.60\times10^{-19}\ \mathrm{C}$. )

Q4. A photon of wavelength $\lambda=500\ \mathrm{nm}$ has momentum $p=h/\lambda$. An electron is prepared so that its momentum has the same magnitude $p$. Using $h=6.63\times10^{-34}\ \mathrm{Js}$ and $m_e=9.11\times10^{-31}\ \mathrm{kg}$, the kinetic energy $K$ of this electron (non-relativistic) is closest to:

Q5. An electron and a proton have identical de Broglie wavelength $\lambda$. If the electron's kinetic energy is $100\ \mathrm{eV}$ and the proton mass is approximately $1836$ times the electron mass, the kinetic energy of the proton is approximately:

Q6. A monochromatic beam of light of wavelength $400\,\mathrm{nm}$ falls on a metal whose work function is $\phi=2.00\,\mathrm{eV}$. Photoelectrons are emitted and the stopping potential $V_0$ is measured. (Use $hc=1240\,\mathrm{eV\cdot nm}$ and $K_{\max}=h\nu-\phi$.) The stopping potential $V_0$ is approximately:

Q7. A photon of wavelength $0.010\,\mathrm{nm}$ is Compton scattered by a free electron and is back-scattered ($\theta=180^\circ$). Using $hc=1240\,\mathrm{eV\cdot nm}$ and the Compton wavelength $\lambda_C=\dfrac{h}{m_e c}=0.00243\,\mathrm{nm}$, the maximum kinetic energy transferred to the recoil electron is closest to:

Q8. To resolve two features separated by $0.10\,\mathrm{nm}$ in an electron microscope one needs electrons with de Broglie wavelength of order this separation or smaller. Using the non-relativistic approximation $\lambda(\mathrm{nm})\approx\dfrac{1.226}{\sqrt{V(\mathrm{V})}}$, the minimum accelerating potential $V$ required is approximately:

Q9. Assertion: If one attempts to determine which slit an electron passes through in a double-slit experiment by illuminating the slits with light of wavelength $\lambda$ comparable to the slit separation $d$, the interference pattern disappears.
Reason: The interference is destroyed because the electron absorbs the illuminating photon and its de Broglie wavelength becomes equal to the photon's wavelength, removing the phase correlation required for interference.

Q10. For Compton scattering the maximum kinetic energy transferred to the electron (back-scattering, $\theta=180^\circ$) can be written as $K_{\max}=\dfrac{2E^2}{m_e c^2+2E}$ where $E$ is the incident photon energy. For what incident photon energy $E$ does $K_{\max}$ equal the rest energy $m_e c^2$ of the electron? (Take $m_e c^2=511\,\mathrm{keV}$.) The closest value of $E$ is:

Q11. An electron is accelerated through a potential difference of $150\ \text{V}$. Assuming non-relativistic motion and using $\displaystyle \lambda=\frac{h}{\sqrt{2m_e eV}}$, its de Broglie wavelength $\lambda$ is approximately:

Q12. An electron beam accelerated through $200\ \text{V}$ is diffracted by a crystal and gives a first-order Bragg peak ($n=1$) at angle $\theta=30^\circ$. Using $\displaystyle \lambda=\frac{h}{\sqrt{2m_e eV}}$ and Bragg's law $n\lambda=2d\sin\theta$, the interplanar spacing $d$ is:

Q13. An X-ray photon of wavelength $0.05\ \mathrm{nm}$ is back-scattered ($\theta=180^\circ$) by a free electron. Using Compton formula $\Delta\lambda=\lambda_C(1-\cos\theta)$ with $\lambda_C=\dfrac{h}{m_e c}=2.426\times10^{-12}\ \mathrm{m}$, the kinetic energy gained by the electron is approximately:

Q14. An electron has de Broglie wavelength $\lambda=0.01\ \mathrm{nm}$. Taking $p=\dfrac{h}{\lambda}$ and using the relativistic relation $E=\sqrt{(pc)^2+(m_e c^2)^2}$, the minimum accelerating potential $V$ (in kV) required to produce such electrons is closest to:

Q15. In an electron diffraction experiment a first-order maximum ($n=1$) appears at $\theta=30^\circ$ when electrons are accelerated through $V=30\ \text{kV}$. A student computes the interplanar spacing $d_{\text{nr}}$ using the non-relativistic de Broglie formula $\lambda_{\text{nr}}=\dfrac{h}{\sqrt{2m_e eV}}$, while the correct relativistic spacing is $d_{\text{r}}$. The percentage error $\dfrac{d_{\text{nr}}-d_{\text{r}}}{d_{\text{r}}}\times100\%$ is approximately: