What is the work function of the metal if the light of wavelength 4000 Å generates photoelectrons of velocity 6105 ms-1…

🧠 The Energy Subtraction A $4000 \text{ \AA}$ photon brings a specific amount of "money" (energy). The electron spends part of it as its "exit fee" (Work Function) and keeps the rest as speed. We just need to work backward from the speed to find the fee.

🗺️ The Work Function Path

1. Incident Photon Energy ($E$): $E = \frac{12400}{4000} = 3.1 \text{ eV}$
2. Kinetic Energy ($KE$): For $v = 6 \times 10^5 \text{ m/s}$: $KE = \frac{1}{2} m v^2 = \frac{1}{2} (9 \times 10^{-31}) (36 \times 10^{10}) = 1.62 \times 10^{-19} \text{ J}$ Convert to eV: $\frac{1.62 \times 10^{-19}}{1.6 \times 10^{-19}} \approx 1.01 \text{ eV}$.
3. Subtracting for $W_0$: $W_0 = E - KE = 3.1 - 1.01 = 2.09 \approx 2.1 \text{ eV}$

⚡ The $12400$ and $KE$ Check A speed of $6 \times 10^5 \text{ m/s}$ roughly corresponds to $1 \text{ eV}$. If you get $10 \text{ eV}$ or $0.1 \text{ eV}$, you've likely missed a power of ten in the Joule-to-eV conversion.

⚠️ Common Traps Using $1242$ instead of $12400$ for Angstroms, or forgetting to square the velocity. $6^2 = 36$, not $12$!

$\boxed{\text{Answer: (d)}}$