Question

The work function of aluminum is 4.2eV. Light of wavelength $2000\mathring{A}$ is incident on it. The minimum energy of emitted electrons will be

• A. Zero
• B. 2.0 eV
• C. 4.2 eV
• D. 6.19 eV

Answer 1

Answer: (B)

2.0 eV

Energy of the incident light is $6.2eV$. Now the work function of aluminum is $4.2$. So the minimum energy of emitted electrons will be $(6.2-4.2)=2$ eV.

So, the answer is option (B).