Question

Radiation of wavelength $3000\stackrel{o}{A}$ falls on a photoelectric surface for which work function is 1.6 eV. What is the stopping potential for emitted electron

• A. 2.53 V
• B. 3.32 V
• C. 1.20 V
• D. 5.20 V

Answer 1

The correct option is A 2.53 V

Given,
Wavelength of radiation = $\left(\lambda \right)=3000\stackrel{o}{A}$
Workfunction = $\varphi =1.6eV$
According to Einstein's photoelectric equation,
$K.E_{max}=h\nu -\varphi =\frac{hc}{\lambda }-\varphi$
Energy of photon of the incident radiation
$E\left(eV\right)=\frac{12400}{\lambda \left(A^0\right)}$
$E\left(eV\right)=\frac{12400}{3000}=4.13eV$
$K.E_{max}=h\nu -\varphi =4.13-1.6=2.53eV$
Therefore the stopping potential of the emitted electron is 2.53 V