The threshold wavelength for the photoelectric effect on sodium is $5000\;\mathring{A}$. Its work function is:

Let $K_1$ be the maximum kinetic energy of photoelectrons emitted by a light of wavelength $\lambda_1$ and $K_2$ corresponding to $\lambda_2$. If $\lambda_1 = 2\lambda_2$,then:

In a photoelectric effect experiment,the threshold wavelength of incident light is $380 \, nm$. If the wavelength of the incident light is $260 \, nm$,the maximum kinetic energy of emitted electrons will be: .............. $eV$. Given $E \text{ (in } eV) = \frac{1237}{\lambda \text{ (in } nm)}$.

Which one of the following four graphs showing lines $P, Q, R$ and $S$ between maximum kinetic energy $(E)$ and intensity of incident light $(I)$ is correct?

If the work function of $Na$ metal is $2.3 \ eV$,in which of the following regions of the $EM$ spectrum does the threshold wavelength lie?

$A$ point source of light is used in a photoelectric effect. If the source is moved farther from the emitting metal, the stopping potential: