Question

Which of the following statements are correct, if the threshold frequency of caesium is $ 5.16 \times 10^{14} \, \text{Hz} $?

• A. When Cs is placed inside a vacuum chamber with an ammeter connected to it and yellow light is focused on Cs, the ammeter shows the presence of current.
• B. When the brightness of the yellow light is dimmed, the value of the current in the ammeter is reduced.
• C. When a red light is used instead of the yellow light, the current produced is higher with respect to the yellow light.
• D. When a blue light is used, the ammeter shows the formation of current.

Hint:

In the photoelectric effect, only light with a frequency greater than the threshold frequency can cause the ejection of electrons. Reducing light intensity reduces the current, but light with insufficient frequency (like red light) will not cause any emission of electrons.

Answer 1

The Correct Option is D

In the photoelectric effect, electrons are ejected from a material when light of a frequency greater than the threshold frequency strikes it. 
Option A: When yellow light (which has a frequency greater than the threshold frequency) is focused on caesium, electrons are ejected, and current flows. Therefore, this statement is correct. 
Option B: Dimming the brightness of the yellow light reduces the number of photons, which in turn decreases the number of electrons ejected. Hence, the current in the ammeter is reduced. This statement is correct. 
Option C: Red light has a frequency lower than the threshold frequency of caesium, so it does not have enough energy to eject electrons. Hence, no current will be produced. This statement is incorrect. 
Option D: Blue light has a frequency greater than the threshold frequency, so it will eject electrons and form current in the ammeter. This statement is correct. 
Thus, the correct answer is B, C, and D Only.

Answer 2

We are given the following formula for calculating wavelength:

λ = c / v

Where:

• λ is the wavelength of the light.
• c is the speed of light in a vacuum, which is approximately 3 × 10⁸ m/s.
• v is the frequency of the light, given as 5.16 × 10¹⁴ Hz in this case.

Substituting the values into the formula:

λ = (3 × 10⁸) / (5.16 × 10¹⁴)

Now, performing the calculation:

λ = 581.39 nm

The wavelength of light is 581.39 nm (nanometers), which falls in the visible light spectrum, specifically near yellow light.

Analysis of the Photoelectric Effect:

• λ_Photon is near and below yellow light. Since the photon energy is related to the frequency and wavelength, photons with a wavelength near or shorter than yellow light can cause the photoelectric effect, which is the phenomenon where electrons are ejected from a material when it is exposed to light.
• If the intensity of the light decreases, the photocurrent will decrease. This is because the intensity of light is proportional to the number of photons hitting the material. Fewer photons will result in fewer electrons being ejected, thus a lower photocurrent.
• Red light will not produce the photoelectric effect because the energy of red photons is too low. The energy of a photon is inversely proportional to its wavelength (higher wavelength means lower energy), and for the photoelectric effect to occur, the photon energy must be greater than the work function of the material (which is typically lower than that of red light).
• In this case, V_Blue > V_Yellow, so a greater voltage is applied with blue light, and thus, more electrons are ejected, creating a higher photocurrent compared to yellow light.
• White light contains all frequencies of visible light, including those that are high enough to produce the photoelectric effect (like blue or ultraviolet light). Therefore, white light will indeed cause a photoelectric current.

Thus, based on the wavelength and the behavior of the photoelectric effect with different light intensities and frequencies, we can confirm that:

• Yellow light is capable of producing the photoelectric effect when its frequency is sufficiently high (as determined by the wavelength).
• Red light, with a lower frequency, will not produce the photoelectric effect.
• White light, containing various frequencies, will cause a photoelectric current as it includes frequencies that are capable of inducing the effect.