Q.Explain the inverse linear dependence of stopping potential on the incident wavelength in a photoelectric effect experiment.

[!TLDR] From Einstein's equation $eV_0=h\nu-\phi_0=\frac{hc}{\lambda}-\phi_0$, so $V_0=\frac{hc}{e\lambda}-\frac{\phi_0}{e}$ -- $V_0$ varies inversely (and linearly in $1/\lambda$) with wavelength: as $\lambda$ increases, $1/\lambda$ (and hence $V_0$) decreases linearly, until $V_0$ reaches zero at the threshold wavelength. [!ANSWER] $V_0=\frac{hc}{e\lambda}-\frac{\phi_0}{e}$: a straight line in $V_0$ vs $1/\lambda$, so $V_0$ falls off inversely as $\lambda$ grows.

Einstein's photoelectric equation, in terms of stopping potential, is $eV_0=KE_{max}=h\nu-\phi_0$. Since frequency and wavelength are related by $\nu=c/\lambda$, substituting gives\n\n$eV_0 = \frac{hc}{\lambda} - \phi_0 \qquad\Longrightarrow\qquad V_0 = \frac{hc}{e\lambda} - \frac{\phi_0}{e}$\n\nThis shows that $V_0$ is NOT linear in $\lambda$ itself, but is linear in $\boldsymbol{1/\lambda}$: plotting $V_0$ against $1/\lambda$ gives a straight line of slope $hc/e$ (a universal constant, same for every metal) and intercept $-\phi_0/e$ (which depends on the emitter material). As the wavelength $\lambda$ INCREASES, $1/\lambda$ decreases, and since $V_0$ is directly proportional to $1/\lambda$ (apart from the constant offset $-\phi_0/e$), $V_0$ correspondingly DECREASES -- an inverse relationship between $V_0$ and $\lambda$, even though the underlying mathematical dependence is a straight line when plotted against $1/\lambda$ rather than against $\lambda$ directly.\n\nPhysically, this makes sense: a longer wavelength means lower frequency, hence a lower-energy photon $h\nu=hc/\lambda$; since $KE_{max}=h\nu-\phi_0$, a lower photon energy directly means a lower maximum kinetic energy for the photoelectrons, and hence a lower stopping potential is needed to turn them back. As $\lambda$ keeps increasing, $V_0$ keeps falling until it reaches exactly zero at the THRESHOLD WAVELENGTH $\lambda_0=hc/\phi_0$ (equivalently, the threshold frequency $\nu_0=\phi_0/h$) -- beyond which no photoelectrons are emitted at all, however long the wavelength is stretched. [!ANSWER] $V_0=\frac{hc}{e\lambda}-\frac{\phi_0}{e}$: a straight line in $V_0$ vs $1/\lambda$, so $V_0$ falls off inversely as $\lambda$ grows.

Substitute $\nu=c/\lambda$ into Einstein's photoelectric equation $eV_0=h\nu-\phi_0$ to get $V_0$ as a function of $1/\lambda$, and interpret the resulting inverse relationship.

Describing the relationship as $V_0$ varying linearly with $\lambda$ itself, rather than with $1/\lambda$ -- the true dependence is an inverse (hyperbolic-looking) relationship in $\lambda$, even though it is exactly linear when plotted against $1/\lambda$ or against frequency $\nu$.