A clean metal surface of work function $\phi_0 = 2.0\,\text{eV}$ is illuminated by monochromatic light of wavelength $\lambda = 400\,\text{nm}$. The maximum kinetic energy of the ejected photoelectrons is closest to (Take $hc = 1240\,\text{eV\,nm}$.)

1. Einstein's photoelectric equation: $K_{\max} = h\nu - \phi_0 = \dfrac{hc}{\lambda} - \phi_0$. 2. Photon energy: $E_{\text{ph}} = \dfrac{hc}{\lambda} = \dfrac{1240\,\text{eV\,nm}}{400\,\text{nm}} = 3.10\,\text{eV}$. 3. Subtract the work function: $K_{\max} = 3.10 - 2.0 = 1.10\,\text{eV}$. 4. Sanity check: the threshold wavelength for this metal is $\lambda_0 = hc/\phi_0 = 1240/2.0 = 620\,\text{nm}$. Since $400 < 620\,\text{nm}$, the incident light is above threshold and we EXPECT a positive $K_{\max}$. ✓ 5. Distractor analysis. Option D ($3.1\,\text{eV}$) is the bare photon energy without subtracting $\phi_0$. Option B ($0.8\,\text{eV}$) uses $hc = 1120$, a wrong constant. Option A ($0.5\,\text{eV}$) halves the result — a sign-management slip. _Source: NCERT Class 12 Physics Part 2, Ch 11 "Dual Nature of Radiation and Matter", §11.6 (Einstein equation, Example 11.2 method), p. 7 + p. 11._