A photon of energy $10\,\text{eV}$ strikes a metal of work function $4\,\text{eV}$. The maximum speed acquired by the emitted photoelectron is closest to (Take $m_e = 9.11\times 10^{-31}\,\text{kg}$ and $1\,\text{eV} = 1.6\times 10^{-19}\,\text{J}$.)

1. Maximum kinetic energy of photoelectron: $K_{\max} = h\nu - \phi_0 = 10 - 4 = 6\,\text{eV}$. 2. Convert to joules: $K_{\max} = 6 \times 1.6\times 10^{-19} = 9.6\times 10^{-19}\,\text{J}$. 3. Relate kinetic energy to speed (non-relativistic, since $v \ll c$): $K_{\max} = \tfrac{1}{2} m_e v_{\max}^2$. 4. Solve: $v_{\max} = \sqrt{\dfrac{2 K_{\max}}{m_e}} = \sqrt{\dfrac{2 \times 9.6\times 10^{-19}}{9.11\times 10^{-31}}}$. 5. Compute the ratio: $\dfrac{1.92\times 10^{-18}}{9.11\times 10^{-31}} = 2.108\times 10^{12}$. 6. Take square root: $v_{\max} = \sqrt{2.108\times 10^{12}} = 1.452\times 10^{6}\,\text{m/s}$, which rounds to $1.45\times 10^{6}\,\text{m/s}$. 7. Distractors: A uses $K_{\max} = 2\,\text{eV}$ (wrong subtraction); C uses $K_{\max} = 12\,\text{eV}$; D is the speed of light (the answer if you forgot the equation entirely). _Source: NCERT Class 12 Physics Part 2, Ch 11, §11.6 (Einstein equation applied to photoelectron speed), p. 7._