Question

If the work function of potassium is $ 2\,eV, $ then its photoelectric threshold wavelength is

• A. 310 nm
• B. 620 nm
• C. 6200 nm
• D. 3100 nm

Answer 1

Answer: (B)

The minimum energy required for the emission of photoelectrons from a metal is called the work function $(W)$ of that metal, and the corresponding wavelength is the threshold wavelength.
$\therefore W=\frac{h c}{\lambda}$
$\Rightarrow \lambda=\frac{h c}{W}$
Given, $h=6.6 \times 10^{-34} J - s,$
$C=3 \times 10^{8} m / s W=2\, eV$
$=2 \times 1.6 \times 10^{-19} J$
$\therefore \lambda=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2 \times 1.6 \times 10^{-19}}$
$=620 \times 10^{-9} m$
$\Rightarrow \lambda =620\, nm$