$\lambda\geq\lambda_0$
$\lambda\geq2\lambda_0$
$\lambda\leq\lambda_0$
$\lambda<\frac{\lambda_ 0}{2}.$
Explanation:
As the work-function of the metal is
$\frac{\text{hc}}{\lambda_0}$, its threshold wavelength is $\lambda_0.$For photoelectric effect, the wavelength of the incident light should be less than or equal to the threshold wavelength of the metal on which light is incident.
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