Question

If the work function of potassium is $2eV,$ 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.
$\therefore W=\frac{hc}{\lambda }$
$\Rightarrow \lambda =\frac{hc}{W}$
Given, $h=6.6\times 10^{-34}J-s,$
$C=3\times 10^{8}m/sW=2eV$
$=2\times 1.6\times 10^{-19}J$
$\therefore \lambda =\frac{6.6\times 10^{-34}\times 3\times 10^8}{2\times 1.6\times 10^{-19}}$
$=620\times 10^{-9}m$
$\Rightarrow \lambda =620nm$