Physics - Modern Physics Question with Solution | TestHub

According to Einstein's photoelectric equation:

$K_{\max }=\frac{hc}{\lambda }-\Phi$

For wavelength ${\lambda }_1$, the maximum kinetic energy is $\frac{1}{2}m{V}^2$:

$\frac{1}{2}m{V}^2=\frac{hc}{{\lambda }_1}-\Phi$

For wavelength ${\lambda }_2$, the maximum kinetic energy is $\frac{1}{2}m(2V{)}^2=4\left(\frac{1}{2}m{V}^2\right)$:

$4\left(\frac{1}{2}m{V}^2\right)=\frac{hc}{{\lambda }_2}-\Phi$

Substituting the value of $\frac{1}{2}m{V}^2$ from the first equation into the second equation:

$4\left(\frac{hc}{{\lambda }_1}-\Phi \right)=\frac{hc}{{\lambda }_2}-\Phi$

$\frac{4hc}{{\lambda }_1}-4\Phi =\frac{hc}{{\lambda }_2}-\Phi$

$3\Phi =\frac{4hc}{{\lambda }_1}-\frac{hc}{{\lambda }_2}$

$3\Phi =hc\left(\frac{4{\lambda }_2-{\lambda }_1}{{\lambda }_1{\lambda }_2}\right)$

$\Phi =\frac{hc}{3{\lambda }_1{\lambda }_2}(4{\lambda }_2-{\lambda }_1)$

Answer: $\frac{\mathrm{hc}}{3{\lambda }_1{\lambda }_2}\left(4{\lambda }_2-{\lambda }_1\right)$