The work function of caesium is $\phi_0 = 2.14\,\text{eV}$. The threshold frequency for photoelectric emission from caesium is closest to which value? (Take $h = 6.63\times 10^{-34}\,\text{J\,s}$ and $1\,\text{eV} = 1.602\times 10^{-19}\,\text{J}$.)

1. At the threshold the photon just supplies the work function, so $h\nu_0 = \phi_0$ and $\nu_0 = \phi_0 / h$. 2. Convert $\phi_0$ to joules: $\phi_0 = (2.14)(1.602\times 10^{-19}) = 3.43\times 10^{-19}\,\text{J}$. 3. $\nu_0 = \dfrac{3.43\times 10^{-19}}{6.63\times 10^{-34}} = 5.16\times 10^{14}\,\text{Hz}$. 4. Rounded this is $5.2\times 10^{14}\,\text{Hz}$. 5. Option B uses $h$ without unit conversion of $\phi_0$; option A halves the result by mistake; option D uses $1\,\text{eV} = 1.0\times 10^{-19}\,\text{J}$, the rounded value. _Source: NCERT Class 12 Physics Part 2, Ch 11, Example 11.2 (part a), p. 11._