Question

Ultraviolet light of wavelength $350\, nm$ and intensity $1.00\, W / m ^{2}$ is directed at a potassium surface. What will be the maximum kinetic energy of the photoelectrons? (Work function of potassium $=2.2\, eV$ )

• A. 3.5 eV
• B. 2.6 eV
• C. 2.2 eV
• D. 1.3 eV

Answer 1

Answer: (D)

Given, $\lambda=350\, nm$
$ =350 \times 10^{-7} m$
Intensity $=1.00\, \omega / m ^{2}$
work function $W =2.2\, eV$
Energy of the photon $=\frac{h c}{\lambda}$
$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{350 \times 10^{8}}$
$=3.5\, eV K_{\max }=E-W$
$=(3.5-2.2) eV =1.3\, eV$