Since, the striking photon has energy equal to hvv and the minimum energy required to eject the electron is hv0 (work function W0), the difference in energy (hv−hv0) is transferred as the kinetic energy of the photoelectron. hv=hv0+KE hv0= minimum energy required to remove electron) Energy of a 300 nm photon is given by ⇒E=λhc=300×10−9 m6.626×10−34Js×3×108 ms−1=6.626×10−19 J Energy of one mole of photons =6.626×10−19 J×6.022×1023 mol−1=3.99×105 J mol−1 The minimum energy needed to remove one mole of electrons =(3.99−1.68)×105 J mol−1 =2.31×105 J mol−1
When electromagnetic radiation of wavelength 300 ~nm falls on the surface of a metal, electrons are emitted with the kinetic energy of 1.68 × 10^5 J…
NEET UG 2022 — Chemistry Physical Chemistry
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 Jmol−1. What is the minimum energy needed to remove an electron from the metal? (h=6.626×10−34Js,c=3×108 ms−1,NA=6.022×1023 mol−1
Held on 30 Apr 2022 · Verified 9 Jul 2026.
Options
- A
2.31×106 J mol−1
- B
3.84×104 J mol−1
- C
3.84×10−19 J mol−1
- D
2.31×105 J mol−1
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