b
Let \(\phi_{0}\) be the work function of the surface of the material. Then, According to Einstein's photoelectric equation, the maximum kinetic energy of the emitted photoelectrons in the first case is

\(K_{\max _{1}}=\frac{h c}{\lambda}-\phi_{0}\)

and that in the second case is

\(K_{\max _{2}}=\frac{h c}{\frac{\lambda}{2}}-\phi_{0}=\frac{2 h c}{\lambda}-\phi_{0}\)

But \(K_{\max _{2}}=3 K_{\max _{1}}(\text { given })\)

\(\therefore \quad \frac{2 h c}{\lambda}-\phi_{0}=3\left(\frac{h c}{\lambda}-\phi_{0}\right)\)

\({\frac{2 h c}{\lambda}-\phi_{0}=\frac{3 h c}{\lambda}-3 \phi_{0}}\)

\({3 \phi_{0}-\phi_{0}=\frac{3 h c}{\lambda}-\frac{2 h c}{\lambda}}\)

\({2 \phi_{0}=\frac{h c}{\lambda} \quad \text { or } \quad \phi_{0}=\frac{h c}{2 \lambda}}\)