The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

Question

The light of two different frequencies whose photons have energies 3.8 eV and 1.4 eV respectively, illuminate a metallic surface whose work function is 0.6 eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be : (A) 1: 1 (B) 2: 1 (C) 4: 1 (D) 1: 4

Tardigrade Admin · Accepted Answer

√(3.8-0.6/1.4-0.6)=√(3.2/0.8)=2

Q. The light of two different frequencies whose photons have energies $3.8 e V$ and $1.4 e V$ respectively, illuminate a metallic surface whose work function is $0.6 e V$ successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

$1 : 1$

$2 : 1$

$4 : 1$

$1 : 4$

$\frac{3.8 - 0.6}{1.4 - 0.6} = \frac{3.2}{0.8} = 2$

Q. The light of two different frequencies whose photons have energies 3.8eV and 1.4eV respectively, illuminate a metallic surface whose work function is 0.6eV successively. The ratio of maximum speeds of emitted electrons for the two frequencies respectively will be :

1:1

2:1

4:1

1:4