Einstein's Photoelectric Equation And Stopping Potential

This problem chains two NCERT relations. First, the photon energy is $E=\frac{hc}{\lambda }=\frac{1242\text{eVcdotpnm}}{400\text{nm}}\approx 3.1$ eV, using the convenient constant $hc=1242$ eV·nm. Next, Einstein's equation gives the maximum kinetic energy as $K_{max}=E-\varphi =3.1-2.0=1.1$ eV. Since the stopping potential satisfies $e{V}_0=K_{max}$, its numerical value in volts equals the kinetic energy in electron-volts, so $V_0\approx 1.1$ V. The value 3.1 V is wrong because it equates the stopping potential with the full photon energy, forgetting to subtract the work function. The value 2.0 V is wrong because it mistakes the work function itself for the stopping potential. The value 0.5 V is wrong because it uses an incorrect photon energy, perhaps from a wavelength error. As stated in NCERT Class 12, Chapter 11, the stopping potential reflects only the surplus energy after the work function is paid. This two-step reasoning, converting wavelength to photon energy and then applying energy conservation, is the standard template for photoelectric numericals. It highlights that the stopping potential is not an arbitrary voltage but a direct experimental readout of the electron's maximum kinetic energy after paying the work-function cost. A magnitude check confirms that a 3.1 eV photon exceeding a 2.0 eV barrier leaves about 1.1 eV, giving a physically sensible stopping potential just above one volt.