Question

For intensity $I$ of a light of wavelength $5000\overset{^\circ }{A}$ the photoelectron saturation is $0.40\mu A$ and stopping potential is $1.36V$ , the work function of metal is

• A. $2.47eV$
• B. $0.36eV$
• C. $1.10eV$
• D. $0.43eV$

Answer 1

Answer: (C)

By using $E=W_{0}+K_{max}$
$E=\frac{12375}{5000}=2.475eV$ and $K_{max}=eV_{0}=1.36eV$
So $2.475=W_{0}+1.36\Rightarrow W_{0}=1.1eV$