A light whose frequency is equal to 6 × 1014 Hz is incident on a metal whose work function is 2 eV .[h=6.63 × 10-34 Js , 1 eV =1.6 × 10-19 J ] The maximum energy of the electrons emitted will be

Question

A light whose frequency is equal to 6 × 1014 Hz is incident on a metal whose work function is 2 eV .[h=6.63 × 10-34 Js , 1 eV =1.6 × 10-19 J ] The maximum energy of the electrons emitted will be (A) 2.49 eV (B) 4.49 eV (C) 0.49 eV (D) 5.49 eV

Tardigrade Admin · Accepted Answer

KE max =h v-φ where h v= energy of incident photon, φ= work function KE max =6.6 × 10-34 × 6 × 1014-2 × 1.6 × 10-19 =3.96 × 10-19-3.2 × 10-19 =(0.76 × 10-19/1.6 × 10-19) eV =0.475 eV

Q. A light whose frequency is equal to 6×1014Hz is incident on a metal whose work function is 2eV.[h=6.63×10−34Js,1eV=1.6×10−19J] The maximum energy of the electrons emitted will be

2.49 eV

4.49 eV

0.49 eV

5.49 eV

KEmax​=hv−ϕ where hv= energy of incident photon, ϕ= work function KEmax​=6.6×10−34×6×1014−2×1.6×10−19 =3.96×10−19−3.2×10−19 =1.6×10−190.76×10−19​eV =0.475eV

Q. A light whose frequency is equal to $6 \times 1 0^{14} Hz$ is incident on a metal whose work function is $2 e V . [h = 6.63 \times 1 0^{- 34} J s, 1 e V = 1.6 \times 1 0^{- 19} J]$ The maximum energy of the electrons emitted will be