Question

Calculate the threshold frequency for the surface.

• A. $1.5\times {10}^{14}Hz$
• B. $3\times {10}^{14}Hz$
• C. $4\times {10}^{14}Hz$
• D. $5\times {10}^{14}Hz$

Answer 1

The correct option is D $5\times {10}^{14}Hz$

Using $e{V}_s=\frac{hc}{\lambda }-\varphi$
where $\varphi$ is the work function of metal, $hc=1241eV$ if wavelength is put in $nm$.
Case 1: $\lambda =550nm$ $V_s=0.19V$
$\therefore$ $0.19eV=\frac{1241}{550}eV-\varphi$
Or $0.19eV=2.256eV-\varphi$
$⟹\varphi =2.07eV$
Threshold frequency for the surface ${\nu }_{th}=\frac{\varphi }{h}=\frac{2.07\times 1.6\times {10}^{-19}}{6.62\times {10}^{-34}}=5\times {10}^{14}Hz$