Physics - Modern Physics Question with Solution | TestHub

Using Einstein's photoelectric equation:

$e{V}_s=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$

For the first case:

$e{V}_1=\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_0}$

For the second case:

$\frac{e{V}_1}{6}=\frac{hc}{3{\lambda }_1}-\frac{hc}{{\lambda }_0}$

Multiplying the second equation by $6$:

$e{V}_1=\frac{6hc}{3{\lambda }_1}-\frac{6hc}{{\lambda }_0}=\frac{2hc}{{\lambda }_1}-\frac{6hc}{{\lambda }_0}$

Equating the two expressions for $e{V}_1$:

$\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_0}=\frac{2hc}{{\lambda }_1}-\frac{6hc}{{\lambda }_0}$

Dividing throughout by $hc$:

$\frac{1}{{\lambda }_1}-\frac{1}{{\lambda }_0}=\frac{2}{{\lambda }_1}-\frac{6}{{\lambda }_0}$

Rearranging the terms:

$\frac{6}{{\lambda }_0}-\frac{1}{{\lambda }_0}=\frac{2}{{\lambda }_1}-\frac{1}{{\lambda }_1}$

$\frac{5}{{\lambda }_0}=\frac{1}{{\lambda }_1}$

${\lambda }_0=5{\lambda }_1$

Answer: $5{\lambda }_1$