Question

The diagram above represents the photoelectric effect for a metal. When the metal surface is exposed to light with increasing frequency and energy of photons, electrons first begin to be ejected from the metal when the energy of the photons is 3.3×10−19J . Which of the following is closest to the frequency of the light with photon energy of 3.3×10−19J ?
5.0×10−53s−1
5.0×10−16s−1
5.0×1014s−1
C5.0×1052s−1

Answer 1

Answer: υ=5.0x10¹⁴s⁻¹

Explanation: Photoelectric Effect is a fenomenon that occur when incident light on a metal surface causes an electron to be ejected. It is possible to determine the number of eletrons and its kinetic energy by measuring the light's intensity and frequency.

According to Einstein, energy carried by light can be calculated as:

$E=h.\upsilon$

in which

h is Planck's constant, with magnitude h=6.626x10⁻³⁴J.s;

υ is frequency of radiation

Then, frequency of the light is

$\upsilon=\frac{E}{h}$

$\upsilon = \frac{3.3.10^{-19}}{6.626.10^{-34}}$

υ ≈ 5x10¹⁴

The frequency of the light with energy of 3.3x10⁻¹⁹J is 5x10¹⁴s⁻¹