In an experiment to study the photoelectric effect,the observed variation of stopping potential $(V_0)$ with the frequency $(\nu)$ of incident radiation is as shown in the figure. The slope and $y$-intercept are respectively

In an experiment with the photoelectric effect,the stopping potential:

Light strikes a metal surface causing photoelectric emission. The wavelength of incident light is $248 \, nm$. If the stopping potential for the ejected electrons is $2.8 \, eV$, then the work function of the metal is (Take $hc = 1240 \, eV \cdot nm$). (in $ \, eV$)

Sodium and copper have work functions $2.3 \ eV$ and $4.5 \ eV$ respectively. Then the ratio of their threshold wavelengths is nearest to

The work function of caesium metal is $2.14 \; eV$. When light of frequency $6 \times 10^{14} \; Hz$ is incident on the metal surface,photoemission of electrons occurs. What is the
$(a)$ maximum kinetic energy of the emitted electrons,
$(b)$ stopping potential,and
$(c)$ maximum speed of the emitted photoelectrons?

$A$ photoelectric material having work-function $\phi_0$ is illuminated with light of wavelength $\lambda$ (where $\lambda < \frac{hc}{\phi_0}$). The fastest photoelectron has a de Broglie wavelength $\lambda_d$. $A$ change in wavelength of the incident light by $\Delta \lambda$ results in a change $\Delta \lambda_d$ in $\lambda_d$. Then the ratio $\frac{\Delta \lambda_d}{\Delta \lambda}$ is proportional to: