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JEE MainPhysicsDual Nature of Matter

The results of a photoelectric effect experiment on an unknown metal surface are recorded in the table below. Incident Wavelength ( ) Stopping Potential ( V_s ) 310 nm 2.0 V 400 nm 1.1 V Based on this data, the threshold wavelength of the metal is: (Take hc = 1240 eV nm )

Options

  1. A207 nm
  2. B620 nm
  3. C90 nm
  4. D2 nm

Correct answer

B. 620 nm

Step-by-step solution

According to Einstein's photoelectric equation: eV_s = hc - Using the first data point ( = 310 nm , V_s = 2.0 V ): 2.0 eV = 1240 310 - 2.0 eV = 4.0 eV - = 2.0 eV We can verify this with the second data point ( = 400 nm , V_s = 1.1 V ): 1.1 eV = 1240 400 - 1.1 eV = 3.1 eV - = 2.0 eV The threshold wavelength ₀ is given by: = hc ₀ ₀ = 1240 2.0 = 620 nm Answer: 620 nm

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