The stopping potential in an experiment on photoelectric is 1.5V. What is the maximum kinetic energy of the photoelectrons emitted? Calculate in Joules.

- Hint: The photoelectric effect is the emission of electrons when an electromagnetic wave, like light, hits a cloth. Electrons emitted in this manner are called photoelectrons. This phenomenon is commonly studied in electronic physics and in fields of chemistry such as quantum chemistry and electrochemistry.

Complete step-by-step answer:
Now from the question, we have
Photoelectric cut-off voltage, ${V_0}$ = 1.5V
K.E. = eV
Where, e = charge of an electron = $1.6 \times {10^{ - 19}}$C
So,
K.E. = $1.6 \times {10^{ - 19}} \times 1.5$
K.E. = $2.4 \times {10^{ - 19}}$J
This is the required answer.

Note: Explanation of photoelectric effect
According to classical electromagnetic theory, the photoelectric effect can be attributed to the transfer of energy from the light to an electron. From this perspective, an alteration in the intensity of light induces changes in the kinetic energy of the electrons emitted from the metal. According to this theory, a sufficiently dim light is predicted to point out a delay between the initial shining of its light and therefore the subsequent emission of an electron.
But the experimental results did not correlate with either of two predictions made by classical theory. Instead, experiments showed that electrons are dislodged by the impingement of light only when it reached or exceeded a threshold frequency. Below the threshold, no electrons are emitted from the material, regardless of the light intensity or the length of time of exposure to the light. When the photoelectron is emitted into a solid rather into a vacuum, the term internal photoemission is often used, and emission into a vacuum distinguished as external photoemission.