Q.Photocurrent recorded in the micro ammeter in an experimental set-up of photoelectric effect vanishes when the retarding potential is more than 0.8 V if the wavelength of incident radiation is 4950 Å. If the source of incident radiation is changed, the stopping potential turns out to be 1.2 V. Find the work function of the cathode material and the wavelength of the second source.

[!TLDR] Using $eV_0=hc/\lambda-\phi_0$ for both sources: $\phi_0=\frac{hc}{\lambda_1}-eV_{0,1}\approx1.71$ eV, then solving for the second wavelength from $eV_{0,2}=\frac{hc}{\lambda_2}-\phi_0$ gives …

For the first source, $\lambda_1=4950$ Å, retarding (stopping) potential $V_{0,1}=0.8$ V. Using $hc=6.63\times10^{-34}\times3\times10^8=1.989\times10^{-25}$ J m $=12431$ eV Å (a convenient constant, since $hc/e=12431$ eV Å when $\lambda$ is measured in angstroms):\n\n$\frac{hc}{\lambda_1} = \frac{12431}{4950} \approx 2.511\text{ eV}$\n\nFrom Einstein's equation $eV_0=\dfrac{hc}{\lambda}-\phi_0$ (in eV, since $eV_0$ numerically equals $V_0$ in volts):\n\n$\phi_0 = \frac{hc}{\lambda_1} - V_{0,1} = 2.511 - 0.8 = 1.711\text{ eV} \approx 1.71\text{ eV}$\n\nFor the SECOND source, the stopping potential is $V_{0,2}=1.2$ V, using the SAME cathode (same $\phi_0$). Rearranging Einstein's equation for the wavelength:\n\n$\frac{hc}{\lambda_2} = V_{0,2} + \phi_0 = 1.2 + 1.711 = 2.911\text{ eV}$\n\n$$\lambda_2 = \frac{12431}{2.911} \ …

Apply Einstein's equation $eV_0=hc/\lambda-\phi_0$ twice -- once to find $\phi_0$ from the first (known-wavelength) source, then again, rearranged …

Using two different, inconsistent forms of $hc/e$ (e.g. mixing an approximate 12400 eV Å with the more precise 12431 eV Å) across the two parts of the calc …