Light of wavelength $4000{\text{A}}^{\circ }$  is incident on a potassium surface whose work function is $2.1\text{ev}$ . What will be the stopping potential

Wavelength $\text{λ}=4000{\text{A}}^{\circ }$

Work function $w=2.1\text{ev}$

$E=w+{\text{ev}}_{\text{0}}$

$\frac{12,400}{4000}=2.1+e{v}_0$ $\left(\because E=\frac{12400}{\lambda }{\text{A}}^{\circ }\right)$

$3.1=2.1+e{v}_0$

  $v_0=1\text{v}$