Modern Physics — Easy JEE physics MCQ
The work function of a metal is $4.2$ eV. The maximum wavelength of light that can eject photoelectrons from this metal is (Take $hc = 1240$ eV·nm):
- A. $295$ nm
- B. $310$ nm
- C. $350$ nm
- D. $400$ nm
Solution
For photoelectric emission to occur, the photon energy must be at least equal to the work function.
At threshold (maximum wavelength):
$$E = \phi$$
$$\frac{hc}{\lambda_{max}} = \phi$$
$$\lambda_{max} = \frac{hc}{\phi} = \frac{1240}{4.2} = 295.2 \text{ nm}$$