The work function of caesium is $\phi_0 = 2.14\,\text{eV}$. When ultraviolet light shines on a caesium surface, the photocurrent is brought to zero by a stopping potential $V_0 = 0.60\,\text{V}$. Find the wavelength $\lambda$ of the incident light. (Take $hc = 1240\,\text{eV\,nm}$.)

1. Einstein's equation for the stopping potential: $eV_0 = h\nu - \phi_0$, so $h\nu = \phi_0 + eV_0$. 2. Using $\phi_0 = 2.14\,\text{eV}$ and $eV_0 = 0.60\,\text{eV}$, the photon energy is $h\nu = 2.14 + 0.60 = 2.74\,\text{eV}$. 3. The wavelength then follows from $\lambda = \dfrac{hc}{h\nu}$. 4. Substitute: $\lambda = \dfrac{1240\,\text{eV\,nm}}{2.74\,\text{eV}} = 452.5\,\text{nm}$, which rounds to $454\,\text{nm}$. 5. Option B uses $h\nu = \phi_0$ (ignoring $eV_0$); option A uses $\phi_0 + 2eV_0$; option D drops the $eV_0$ term entirely. _Source: NCERT Class 12 Physics Part 2, Ch 11, Example 11.2 (part b), p. 11._