The work functions for metals $$A,B$$  and $$C$$ are respectively $$1.92\,eV,\,2.0\,eV$$   and $$5\,eV.$$  According to Einstein’s equation, the metals which will emit photoelectrons for a radiation of wavelength $$4100\,\mathop {\text{A}}\limits^ \circ $$  is/are

Solution :
Work function for wavelength of $$4100\,\mathop {\text{A}}\limits^ \circ $$  is given by
$$\eqalign{ & {W_0} = \frac{{hc}}{\lambda } = \frac{{6.62 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}{{4100 \times {{10}^{ - 10}}}} \cr & = 4.8 \times {10^{ - 19}}J \cr} $$
Energy in $$eV$$ is given by
$$\eqalign{ & = \frac{{4.8 \times {{10}^{ - 19}}}}{{1.6 \times {{10}^{ - 19}}}}\,eV \cr & = 3\,eV \cr} $$
Now, we have
$$\eqalign{ & {W_A} = 1.92\,eV, \cr & {W_B} = 2.0\,eV, \cr & {W_C} = 5\,eV, \cr} $$
Since, $${W_A} < W$$
and $${W_B} < W,$$  hence, $$A$$ and $$B$$ will emit photoelectrons.