Photoelectric Effect MCQ Quiz

Threshold frequency is the minimum frequency of light required to eject an electron from a metal surface. Similarly, the work function is the minimum energy required to remove an electron from a metal surface. Hertz is a unit of frequency, while Joule is a unit of energy. Therefore, the relationship between hertz and Joule is similar to the relationship between threshold frequency and work function.

Ultraviolet rays are the most effective radiations for the emission of electrons from a metallic surface. This is because ultraviolet rays have higher energy compared to infrared and microwaves, and they can cause the electrons in the metal to gain enough energy to overcome the surface barrier and be emitted. X-rays also have high energy, but ultraviolet rays are specifically mentioned in the question, so they are the correct answer.

Point P on the graph represents the threshold frequency. This is the minimum frequency of radiation required to emit photoelectrons from a photosensitive surface. Below this frequency, no photoelectrons are emitted regardless of the intensity of the radiation. At and above the threshold frequency, the maximum kinetic energy of the emitted photoelectrons increases with increasing frequency of radiation.

The work function of a metal is the minimum amount of energy required to remove an electron from the metal's surface. The maximum kinetic energy of the ejected electrons is equal to the energy of the incident photons minus the work function. In this case, the maximum kinetic energy is given as 1.20 eV. Therefore, the work function of the metal can be calculated by subtracting the maximum kinetic energy from the energy of the incident photons, which is equivalent to the energy of a photon with a wavelength of 350 nm. The answer is 2.4 eV.

When a photon collides with an electron, its wavelength will increase. This is because the collision transfers energy to the electron, causing it to move to a higher energy level or even be ejected from the atom. According to the wave-particle duality of light, an increase in energy corresponds to a decrease in wavelength. Therefore, when a photon transfers energy to an electron, the wavelength of the photon increases.

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When photons with energy equal to or greater than the work function of a material are incident on the material, photoelectrons are emitted. The maximum kinetic energy of the ejected photoelectrons can be calculated by subtracting the work function from the energy of the incident photons. In this case, the energy of the incident photons is 16 eV and the work function is 4.0 eV. Therefore, the maximum kinetic energy of the ejected photoelectrons would be 16 eV - 4.0 eV = 12 eV.

The current produced by photoelectrons is directly proportional to the intensity of the incident light because higher intensity means more photons striking the metal surface, resulting in more photoelectrons being emitted.

When the frequency of the incident light increases from 2X to 4X, the energy of the photons also increases. This means that the photons have enough energy to overcome the threshold frequency and release photoelectrons. As a result, more photoelectrons are emitted, leading to an increased current. Since the frequency has doubled, the energy of the photons has also doubled, and therefore the resulting current of photoelectrons is increased by a factor of 3.

The energy of a photon is directly proportional to its wavelength. The equation E = hc/λ relates energy (E) to Planck's constant (h), the speed of light (c), and wavelength (λ). By rearranging the equation to solve for wavelength, we get λ = hc/E. Plugging in the values given, we find λ = (6.63 x 10^-34 J s)(3.00 x 10^8 m/s)/(3.2 x 10^-19 J) ≈ 2.07 x 10^-7 m = 207 nm. Therefore, the correct answer is 210 nm.