Q.Given the following data for incident wavelength and the stopping potential obtained from an experiment on photoelectric effect, estimate the value of Planck's constant and the work function of the cathode material. What is the threshold frequency and corresponding wavelength? What is the most likely metal used for emitter? Incident wavelength (in Å): 2536, 3650. Stopping potential (in V): 1.95, 0.5.

[!TLDR] From the slope of $V_0$ vs $1/\lambda$ between the two data points: $h\approx6.427\times10^{-34}$ J s and $\phi_0\approx2.801$ eV (matching the book) -- but recomputing the threshold frequency/wavelength from these SAME two numbers gives $\nu_0=\phi_0/h\approx6.97\times10^{14}$ Hz and $\lambda_0\approx4301$ Å, not the printed 6.761×10^14 Hz/4438 Å (an internal inconsistency in the book's own printed key, flagged below). Emitter material: calcium …

Using $eV_0=\dfrac{hc}{\lambda}-\phi_0$, written with $K\equiv hc/e$ (in V·Å, since $\lambda$ is in Å) as the unknown constant to be estimated from the data (not assuming the precise textbook value of h):\n\n$V_0 = \frac{K}{\lambda} - \phi_0 \quad\text{(all energies in eV/volts, } \lambda \text{ in Å)}$\n\nFor the two data points ($\lambda_1=2536$ Å, $V_{0,1}=1.95$ V) and ($\lambda_2=3650$ Å, $V_{0,2}=0.5$ V):\n\n$1.95 = \frac{K}{2536} - \phi_0, \qquad 0.5 = \frac{K}{3650} - \phi_0$\n\nSubtracting: $1.45 = K\left(\dfrac{1}{2536}-\dfrac{1}{3650}\right) = K\times1.2035\times10^{-4}$, so\n\n$K = \frac{1.45}{1.2035\times10^{-4}} \approx 12048\text{ V Å}$\n\nConverting K to Planck's constant: $h = \dfrac{K\times e\times10^{-10}}{c} = \dfrac{12048\times1.6\times10^{-19}\times10^{-10}}{3\times10^8} \approx 6.426\times10^{-34}\text{ J s} \approx 6.427\times10^{-34}\text{ J s}$, matching the book.\n\nThe work function follows from either data point, e.g. $\phi_0=\dfrac{K}{\lambda_2}-V_{0,2}=\dfrac{12048}{3650}-0.5\approx3.301-0.5=2.801$ eV -- also matching the book exactly, and cross-checked with the first point: $\dfrac{12048}{2536}-1.95\approx4.751-1.95=2.801$ eV. Consistent.\n\nNow, the threshold frequency and wavelength are where $V_0=0$: $\lambda_0=K/\phi_0=12048/2.801\approx4301$ Å, and $\nu_0=c/\lambda_0=(3\times10^8)/(4301\times10^{-10})\approx6.974\times10^{14}$ Hz. Equivalently, $\nu_0=\phi_0/h=(2.801\times1.6\times10^{-19})/(6.427\times10^{-34})\approx6.97\times10^{14}$ Hz -- the SAME result, confirming this is the mathematically consistent threshold point for the h and $\phi_0$ values derived above (and matching the book's own h and $\phi_0$).\n\nA flagged discrepancy: the printed textbook answer key states the threshold as $6.761\times10^{14}$ Hz and 4438 Å, but these two numbers are NOT what follows from the SAME h and $\phi_0$ the book itself prints just before them (checking: $\phi_0/h$ …

Use the two-point slope of $V_0$ vs $1/\lambda$ to get $h$ (from the slope $hc/e$) and $\phi_0$ (from either data point), then find the threshold point ($V_0=0$) from those SAME two …

Silently copying a printed answer's threshold frequency/wavelength without checking that it actually follows from the h and work-function values derived earlier in the SAME solution -- here the two do not match, and the internally consistent valu …