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NEET2026PhysicsChapterActual

The given graph shows the variation of stopping potential ( V₀ ) with the frequency ( ) of incident radiation for two different photosensitive metals, M₁ and M₂ . Based on the graph, which of the following statements is correct?

Options

  1. AMetal M₂ has a higher work function than metal M₁ .
  2. BThe slope of the line for M₂ is greater than the slope of the line for M₁ .
  3. CMetal M₁ has a shorter threshold wavelength than metal M₂ .
  4. DFor a given frequency > ₀₂ , the maximum kinetic energy of photoelectrons is higher for M₂ than for M₁ .

Correct answer

A. Metal M₂ has a higher work function than metal M₁ .

Step-by-step solution

According to Einstein's photoelectric equation, eV₀ = h - , where V₀ is the stopping potential, is the frequency of incident radiation, and is the work function of the metal. Rearranging the equation, we get V₀ = ( h e ) - e . This represents a straight line with slope h e and x-intercept ₀ = h , which is the threshold frequency. From the given graph, the threshold frequency for metal M₂ is greater than that for metal M₁ , i.e., ₀₂ > ₀₁ . Since work function = h ₀ , it follows that ₂ > ₁ . Thus, metal M₂ has a high

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