Dual Nature of Radiation and Matter Set-1

Q1. Monochromatic light of wavelength $400\ \text{nm}$ is incident on a metal whose work function is $2.5\ \text{eV}$. The maximum kinetic energy $K_{\max}$ of photoelectrons (in eV) is:

Q2. In a photoelectric experiment the stopping potential $V_s$ versus frequency $f$ is linear. Two measured points are $(f,V_s)=(9.0\times10^{14}\ \text{Hz},\,1.0\ \text{V})$ and $(1.30\times10^{15}\ \text{Hz},\,3.0\ \text{V})$. Using $e=1.60\times10^{-19}\ \text{C}$ determine (i) Planck's constant $h$ (in J·s) and (ii) the work function $\phi$ (in eV) of the metal.

Q3. Assertion (A): For a metal illuminated by monochromatic light with frequency $\nu>\nu_0$, increasing the intensity of the light increases the stopping potential $V_s$.
Reason (R): The stopping potential satisfies $eV_s=K_{\max}=h\nu-\phi$, which depends only on $\nu$ and $\phi$, not on intensity.

Q4. Electrons are accelerated from rest through a potential $V$ and then diffracted by a crystal with interplanar spacing $d=0.20\ \text{nm}$. For first-order Bragg reflection ($n=1$) the Bragg angle is $\theta=30^\circ$. Assuming non-relativistic electrons and $h=6.63\times10^{-34}\ \text{J·s}$, $m=9.11\times10^{-31}\ \text{kg}$, the required accelerating potential $V$ (in volts) is nearest to:

Q5. Two photocurrent $I$ vs applied potential $V$ curves are measured for the same metal under illumination by light of the same frequency (above threshold) but different intensities $I_1$ and $I_2$ where $I_2>I_1$. Which statement correctly describes the two I–V curves?

Q6. An X‑ray photon of wavelength $\lambda=0.0500\ \text{nm}$ is Compton‑scattered by a stationary electron through angle $\theta=60^\circ$. Using $\lambda_C=\dfrac{h}{m_e c}=2.426\times10^{-12}\ \text{m}$ and $hc=1240\ \text{eV·nm}$, the loss of photon energy (initial minus scattered) is approximately:

Q7. Assertion (A): In a single‑electron double‑slit experiment each electron goes through both slits as a matter wave and the interference pattern emerges only after many electrons are detected one by one.
Reason (R): The interference pattern appears after many detections because each electron's de Broglie wave collapses at the detector and only the accumulation of collapsed spots reproduces the interference fringes.

Q8. In an electron diffraction experiment the first‑order maximum is at angle $\theta_1=6.0^\circ$ when electrons are accelerated through $V_1=400\ \text{V}$. If the accelerating voltage is changed to $V_2=100\ \text{V}$ (non‑relativistic), the first‑order angle $\theta_2$ will be approximately:

Q9. Assertion (A): Compton scattering provides direct evidence for the particle nature of light because the scattered photon shows a wavelength shift that depends on scattering angle but not on intensity.
Reason (R): The Compton shift $\Delta\lambda=\lambda_C(1-\cos\theta)$ is derived by conserving energy and momentum while treating the photon as a particle with momentum $p=h/\lambda$, a description that classical wave theory cannot give.

Q10. An electron and a neutron have the same kinetic energy $K=10\ \text{eV}$. Given $m_e=9.11\times10^{-31}\ \text{kg}$ and $m_n=1.675\times10^{-27}\ \text{kg}$, the ratio of their de Broglie wavelengths $\lambda_e/\lambda_n$ is approximately: