The photoelectric cut-off voltage in a certain experiment is $1.5 \ V$. What is the maximum kinetic energy of the emitted photoelectrons?

When a metal surface is illuminated with light of wavelength $\lambda$,the stopping potential is $V$. When the same surface is illuminated by light of wavelength $2\lambda$,the stopping potential is $V/3$. The threshold wavelength for the surface is:

The surface of a certain metal is first illuminated with light of wavelength $\lambda_1 = 350 \ nm$ and then with light of wavelength $\lambda_2 = 540 \ nm$. It is found that the maximum speed of the photoelectrons in the two cases differs by a factor of $2$. The work function of the metal (in $eV$) is close to: (Energy of photon $= \frac{1240}{\lambda \text{ (in } nm)} \ eV$)

$A$ material $A$ exhibits the photoelectric effect. Its work function is $2.5 \ eV$ and the threshold wavelength is $\lambda$. Another material $B$ has a work function of $5 \ eV$. Find the threshold wavelength required to produce the photoelectric effect in $B$.

In the case of the photoelectric effect,the graph of measured stopping potential $(V_0)$ against the frequency $(\nu)$ of incident light is a straight line. The slope of this line multiplied by the charge of an electron $(e)$ gives:

The work function of a substance is $4.0 \,eV$. The longest wavelength of light that can cause photoelectron emission from this substance is approximately ......... $nm$.