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 -1. What is the minimum energy needed to remove an electron from the metal? ( h =6.626 × 10-34 Js , c =3 × 108 ms -1. text , .N A =6.022 × 1023 mol -1)

Question

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 -1. What is the minimum energy needed to remove an electron from the metal? ( h =6.626 × 10-34 Js , c =3 × 108 ms -1. text , .N A =6.022 × 1023 mol -1) (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

Tardigrade Admin · Accepted Answer

Energy of a 300 nm photon is given by ⇒ E =( hc /λ)=(6.626 × 10-34 Js × 3 × 108 ms -1/300 × 10-9 m ) =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

$2.31 \times 1 0^{6} J m o l^{- 1}$

$3.84 \times 1 0^{4} J m o l^{- 1}$

$3.84 \times 1 0^{- 19} J m o l^{- 1}$

$2.31 \times 1 0^{5} J m o l^{- 1}$

Q. When electromagnetic radiation of wavelength 300nm falls on the surface of a metal, electrons are emitted with the kinetic energy of 1.68×105Jmol−1. What is the minimum energy needed to remove an electron from the metal? (h=6.626×10−34Js,c=3×108ms−1, NA​=6.022×1023mol−1)

2.31×106Jmol−1

3.84×104Jmol−1

3.84×10−19Jmol−1

2.31×105Jmol−1

$2.31 \times 1 0^{6} J m o l^{- 1}$

$3.84 \times 1 0^{4} J m o l^{- 1}$

$3.84 \times 1 0^{- 19} J m o l^{- 1}$

$2.31 \times 1 0^{5} J m o l^{- 1}$