Physics - Modern Physics Question with Solution | TestHub

Given,${\text{K}}_{\max }=3\text{eV}$

$h=4.125\times {10}^{-15}\text{eV}{\text{Hz}}^{\text{-1}}$

$h\nu =K_{\max }+W$

$4.125\times {10}^{-15}\times {10}^{15}=3+W$

or$4.125= 3 +W$

or$W=1.125$

The threshold frequency

${\nu }_0=\frac{W}{h}$

${\nu }_0=\frac{1.125}{4.125\times {10}^{-15}}$

${\nu }_0=\frac{1.125\times {10}^{-15}}{4.125}$

${\nu }_0=2.72\times {10}^{14}$