Question

Light of wavelenght ${\lambda }_1$ shines on a metal surface with intensity $x$ and the metal emits $y$ electrons per second of average energy $z$. What will happen to $y$ and $z$, if $x$ is doubled?

• A. $y$ will be doubled and $z$ will become half
• B. $y$ will remian same and $z$ will be doubled
• C. Both $y$ and $z$ will be doubled
• D. $y$ will be doubled but $z$ will remain same

Answer 1

The correct option is D $y$ will be doubled but $z$ will remain same

The photoelectric current or the rate of emission of photoelectrons is directly proportional to the intensity of incident radiation. So, if we double the intensity of light $x$, the photoelectric current or electrons per second $y$ is also doubled.
The negative potential at which photoelectric current becomes zero is called "stopping potential" or "cut-off potential" denoted by $V_o$. Maximum kinetic energy $E_k$ is given by formula:
$E_k=e{V}_o$
So, the maximum kinetic energy does not depend on the intensity of incident light. Therefore, $z$ will remain same.