Given:
The problem involves the relation between the incident frequency and the threshold frequency for the emission of photoelectrons. We are given the following inequality: \[ \frac{f}{2} < f_0 \] where \( f \) is the incident frequency, and \( f_0 \) is the threshold frequency.
Step 1: Understanding the relationship between incident frequency and threshold frequency
The inequality \( \frac{f}{2} < f_0 \) implies that the incident frequency \( f \) is less than twice the threshold frequency \( f_0 \). In this context, it indicates that the incident light has a frequency that is less than the required threshold frequency needed for the emission of photoelectrons.
Step 2: Explanation of the photoelectric effect
In the photoelectric effect, photoelectrons are emitted from the surface of a material when it is illuminated by light of a frequency greater than or equal to the threshold frequency \( f_0 \). If the incident frequency is less than the threshold frequency, the energy of the photons is insufficient to release the photoelectrons.
Step 3: Conclusion from the given condition
Since \( f \) is less than the threshold frequency \( f_0 \), the photons do not possess enough energy to overcome the work function of the material. This leads to the non-emission of photoelectrons.
Conclusion:
Therefore, the current, which depends on the number of photoelectrons emitted, is zero: \[ \Rightarrow \text{current} = 0 \]
Final Answer:
No emission of photoelectrons occurs, leading to no current being generated in this case.
In the photoelectric effect, electrons are emitted only when the frequency of the incident light \( f \) is greater than or equal to the threshold frequency \( f_0 \) of the material. This condition can be written as:
\[ f \geq f_0. \]
Initially, the frequency of the light is \( 1.5 f_0 \), which is above the threshold frequency, so photoelectrons are emitted. However, if the frequency is halved, the new frequency \( f' \) becomes:
\[ f' = \frac{1.5 f_0}{2} = 0.75 f_0. \]
Since \( f' < f_0 \), the incident frequency is now less than the threshold frequency. Therefore, no photoelectrons will be emitted, regardless of the light’s intensity. Emission of photoelectrons depends on frequency, not intensity, when the frequency is below the threshold.
Thus, the answer is: Zero.
Method used for separation of mixture of products (B and C) obtained in the following reaction is: 