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Step-by-Step Solution
Step 1: Identify the Initial Frequency
The light initially has a frequency $ v $ which is given to be $1.5$ times the threshold frequency $v_0$. Mathematically,
$$ v = 1.5\,v_0 = \frac{3}{2}\,v_0. $$
Step 2: Determine the New Frequency
The problem states that the frequency is halved, so the new frequency $ v' $ is
$$ v' = \frac{v}{2} = \frac{1}{2} \left( \frac{3}{2} v_0 \right) = \frac{3}{4}v_0. $$
Step 3: Compare the New Frequency with the Threshold Frequency
The new frequency $v' = \frac{3}{4}v_0$ is less than the threshold frequency $v_0$. For photoemission to occur, the incident light frequency must be at least $v_0$.
Step 4: Recognize the Effect of Lower Frequency and Increased Intensity
Even though the intensity is doubled, if the new frequency is below the threshold frequency, electrons will not be ejected from the material. Intensity only affects how many photons hit the surface per unit time, but if each photon’s energy $ (h v') $ is insufficient to overcome the work function (associated with $v_0$), no photoelectrons are emitted.
Step 5: Conclude the Photoelectric Current
Since $v' < v_0$, no photoelectric emission takes place. Therefore, the photoelectric current is:
Zero.
