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

Answer 1

Answer: (B)

2 : 1

Answer 2

$3.8 = 0.6 + \frac{1}{2} mv^2_1$

$1.4 = 0.6+\frac{1}{2}mv^2_2$

⇒ $\frac{v^2_1}{v^2_2}=\frac{3.2}{0.8}=\frac{4}{1}$

⇒ $\frac{2}{1}$

⇒ $2:1$