In an experiment on photoelectric effect it was observed that for incident light of wavelength $1.98×10^{-7}m$, stopping potential is 2.5 V. What is the work function $\phi_0$ and threshold frequency?

$3.75 eV,\,9.1×10^{14}Hz$

$K_{max}=eV_s=2.5eV$

Where V_s is the incident light

$E=\frac{hc}{λ}=\frac{6.6×10^{-34}×3×10^8}{1.98×10^{-7}}J$

$=\frac{6.6×10^{-34}×3×10^8}{1.98×10^{-7}×1.6×10^{-19}}eV=6.25eV$

According to Einstein’s photoelectric equation

$K_{max}=E-\phi_0$ where $\phi_0$ is the work function

$\phi_0=E-K_{max}=6.25 eV-2.5 eV = 3.75 eV$

$\phi_0=hv_0$

Where v₀ is the threshold frequency

or $v_0=\frac{\phi_0}{h}=\frac{3.7×1.6×10^{-19}}{6.6×10^{-34}}=9.1×10^4Hz$