Physics - Modern Physics Question with Solution | TestHub

According to Einstein photelectric equation

$K.E=\frac{hc}{\lambda }-\varphi$

$\frac{hc}{{\lambda }_1}=K{E}_1+\varphi$.... (i) (for${\lambda }_1)$

$\frac{hc}{{\lambda }_2}=K{E}_2+\varphi$.... (ii) (for${\lambda }_2)$

Let${\lambda }_1=\lambda$and${\lambda }_2=\frac{{\lambda }_1}{2}=\frac{\lambda }{2}$

From Eqs. (i) and (ii),

$\frac{K{E}_2}{K{E}_1}=\frac{\frac{hc}{{\lambda }_2}-\varphi }{\frac{hc}{{\lambda }_1}-\varphi }=\frac{\frac{hc}{\frac{\lambda }{2}}-\varphi }{\frac{hc}{\lambda }-\varphi }=\frac{\frac{2hc}{\lambda }-\varphi }{\frac{hc}{\lambda }-\varphi }$

$=\frac{\frac{hc}{\lambda }-\varphi }{\frac{hc}{\lambda }-\varphi }+\frac{\frac{hc}{\lambda }}{\frac{hc}{\lambda }-\varphi }$

$\because \frac{\frac{hc}{\lambda }}{\frac{hc}{\lambda }-\varphi }>1$

$\therefore \frac{K{E}_2}{K{E}_1}=1+\left(>1\right)>2$

Thus,$K{E}_1<\frac{K{E}_2}{2}$