A) \[1:2\]
B) \[2:1\]
C) \[4:1\]
D) \[1:4\]
Correct Answer: B
Solution :
The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal. Given by \[W=hv\] ...(i) where v is threshold frequency and h is Planck's constant. Also, \[v=\frac{c}{\lambda }\] ?..(ii) where c is speed of light and X the threshold wavelength. From Eqs. (i) and (ii), we get \[W=\frac{hc}{\lambda }\] Given, \[{{\lambda }_{1}}=300nm,\,\,\,\,{{\lambda }_{2}}=600nm.\] \[\therefore \] \[\frac{{{\lambda }_{2}}}{{{\lambda }_{1}}}=\frac{{{W}_{1}}}{{{W}_{2}}}\] \[\Rightarrow \] \[\frac{{{W}_{1}}}{{{W}_{2}}}=\frac{600}{300}=\frac{2}{1}\]
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