Physics - Modern Physics Question with Solution | TestHub

Eienstein's photoelectric equation is given as,

$K_1=hc\left(\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_0}\right)$...(i)

and$K_2=hc\left(\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_0}\right)$...(ii)

Dividing Equation (i) by Equation (ii), we get

$\frac{K_1}{K_2}=\frac{\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_0}}{\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_0}}$

Since,${\lambda }_1=3{\lambda }_2$

$\frac{K_1}{K_2}=\frac{\frac{1}{{3\lambda }_2}-\frac{1}{{\lambda }_0}}{\frac{1}{{\lambda }_2}-\frac{1}{{\lambda }_0}}$

$=\frac{1}{{3\lambda }_2} \frac{\left(1-\frac{3}{\left({\lambda }_0/{\lambda }_2\right)}\right)}{\frac{1}{{\lambda }_2}\left(1-\frac{1}{{\lambda }_0}\right)}$

$\frac{K_1}{K_2}=\frac{1}{3} \frac{\left(1-\frac{3}{\left({\lambda }_0/{\lambda }_2\right)}\right)}{\left(1-\frac{1}{{\lambda }_0/{\lambda }_2}\right)}$

$\frac{{3K}_1}{K_2}=\frac{\left[1-3/\left({\lambda }_0/{\lambda }_2\right)\right]}{\left(1-1/\frac{{\lambda }_0}{{\lambda }_2}\right)}<1$

$\therefore \frac{{3K}_1}{K_2}<1\Rightarrow K_1<\frac{K_2}{3}$