The graph shows variation of stopping potential V0 versus frequency of incident radiation v for two photosensitive metals A and Which one of the two metals has higher threshold frequency and why?

Without access to the graph you mentioned, I can't directly analyze the data to determine which of the two photosensitive metals, A or B, has a higher threshold frequency. However, I can provide a general explanation based on the concept of threshold frequency.

The threshold frequency (f0f0) is the minimum frequency of incident radiation required to eject electrons from the surface of a photosensitive material. Metals with higher work functions typically have higher threshold frequencies.

Here's why:

1. Work Function Relationship: The work function (WW) of a material is directly proportional to its threshold frequency (f0f0) according to the equation W=hf0W=hf0, where hh is Planck's constant. Therefore, a higher work function corresponds to a higher threshold frequency.

2. Ejection of Electrons: For electrons to be emitted from the surface of a metal when exposed to light, the energy of the incident photons must be greater than or equal to the work function (hf0hf0). If the frequency of the incident radiation is lower than the threshold frequency (f0f0), the photons do not possess enough energy to overcome the work function, and no electrons are emitted.

3. Comparison: Based on the concept above, the metal with the higher stopping potential (associated with the higher work function) likely has the higher threshold frequency. This is because it requires higher-energy photons (higher frequency) to eject electrons from its surface compared to the metal with the lower stopping potential.

Therefore, without the specific data from the graph, we can infer that the metal associated with the higher stopping potential (likely metal B) has the higher threshold frequency due to its higher work function.