Question

The photoelectric threshold of a metal surface is $5200$ $\dot{A}$ if light of $4800$ $\dot{A}$ is incident. then

• A. Photoelectrons are liberated
• B. Photoemission does not occur
• C. Photoemission may occur if intensity is large enough
• D. Cannot predict

Answer 1

The correct option is A Photoelectrons are liberated

$KE=\frac{hc}{\lambda }-W=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}=hc(\frac{1}{\lambda }-\frac{1}{{\lambda }_0}){\lambda }_{0}isthresholdwavelength{\lambda }_0=5200A\lambda =4800A{\lambda }_0>\lambda \frac{1}{\lambda }>\frac{1}{{\lambda }_0}KEispositive.Hencephotoelectronsareliberated.$