Question

The photoelectric emission requires a threshold frequency ${\vartheta }_0$ for a certain metal. If incident photons has wavelengths ${\lambda }_1=2200\AA$ and ${\lambda }_2=1900\AA$ produce two photoelectrons with $K{E}_1$ and $K{E}_2$ respectively and if $K{E}_1=2K{E}_2$ then

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

$\frac{hc}{{\lambda }_1}=h{v}_0+K.E.$

$\frac{hc}{{\lambda }_1}=h{v}_0+2K.E$

$K.E.=\frac{hc}{{\lambda }_1}-\frac{hc}{{\lambda }_0}$

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

$2hc[\frac{1}{2200}-\frac{1}{{\lambda }_0}]=hc[\frac{1}{1900}-\frac{1}{{\lambda }_0}]$

$\frac{2}{2200}-\frac{2}{{\lambda }_0}=\frac{1}{1900}-\frac{1}{{\lambda }_0}$

$\frac{1}{{\lambda }_0}=\frac{1}{1100}-\frac{1}{1900}$

$\frac{1}{{\lambda }_0}=\frac{1}{100}\left[\frac{8}{11\times 19}\right]$

${\lambda }_0=100\frac{11\times 9}{8}$

${\lambda }_{0}1237.5\AA$

= $1.2375\times {10}^{-7}m$