Question

A photocathode has a work function of 2.4 ev. the photocathode is illuminated withmonochromatic radiation whose photon energy is 3.5 ev. the wavelength of the illuminatingradiation is closest to:

Answer 1

The photon energy is equal to 
$E=3.5 eV = 5.6 \cdot 10^{-18}J$
We also know that the energy of a photon is equal to
$E=hf$
where h is the Planck constant and f is the photon frequency. By re-arranging this equation, we can find the photon frequency:
$f= \frac{E}{h}= \frac{5.6 \cdot 10^{-18}J}{6.6 \cdot 10^{-34} Js}=8.48 \cdot 10^{15}Hz$

And now we can find its wavelength by using the relationship between frequency, wavelength and a speed of a photon (which is the speed of light, c):
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{8.48 \cdot 10^{15}Hz}=3.54 \cdot 10^{-8} m$