Verified Solution • 2025–26 Session moderate

QNEET-2022-Ch-486: NEET 2022 Chemistry - Class 11 NEET-2022-Ch-486 Solution: When electromagnetic radiation of wavelength 300 nm falls on…

When electromagnetic radiation of wavelength 300 nm falls on the surface of a metal electrons are emitted with the kinetic energy of 1.68×105 J mol−11.68 \times 10^5\ \text{J mol}^{-1}. What is the minimum energy needed to remove an electron from the metal? (h = 6.626×10−34 Js6.626 \times 10^{-34}\ \text{Js} c = 3×108 ms−13 \times 10^8\ \text{ms}^{-1} NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23}\ \text{mol}^{-1}) 2.31×105 J mol−12.31 \times 10^5\ \text{J mol}^{-1} 2.31×106 J mol−12.31 \times 10^6\ \text{J mol}^{-1} 3.84×104 J mol−13.84 \times 10^4\ \text{J mol}^{-1} 3.84×10−19 J mol−13.84 \times 10^{-19}\ \text{J mol}^{-1}

Multiple Choice Options

  • A 2.31 ×\times _ 10 5 J mol −- _ 1
  • B 2.31 ×\times _ 10 6 J mol −- _ 1
  • C 3.84 ×\times _ 10 4 J mol −- _ 1
  • D 3.84 ×\times _ 10 −- _ 19 J mol −- _ 1

Step-by-Step Verified Solution

Energy of one photon = h c _ ( _ = 300 nm) For one mole photons, E = h c _ _ N A E = 6.626 _ 10 _ 34 _ 3 _ 10 8 _ 6.023 _ 10 23 300 _ 10 _ 9 E = 3.99 _ 10 5 J mol _ 1 Kinetic energy = 1.68 _ 10 5 J mol _ 1 W 0 = E _ K . E . = 3.99 _ 10 5 _ 1.68 _ 10 5 = 2.31 _ 10 5 J mol _ 1

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