( ફોટોનની ઊર્જા $ = \frac{{1240}}{{\lambda {\rm{(in\, nm)}}}}\,eV$)
\(\frac{1240}{540}-\phi=(\mathrm{KE})_{11}=x\) ...... \((2)\)
\((1)\,\,-\,\,(2)\)
\(\frac{1240}{350}-\frac{1240}{540}=3.542-2.296=3 x\)
\(1.246=3 x ; \quad x=0.41\)
\(\phi=2.296-0.41=1.886\)