Question

The photoelectric threshold wavelength for potassium (work function being $2\, eV$ ) is :
(Take $h=6.6 \times 10^{-34}\, J - s$ )

• A. 310 nm
• B. 620 nm
• C. 1200 nm
• D. 2100 nm

Answer 1

Answer: (B)

The minimum energy required for the emission of photoelectron from a metal is called the work function of that metal.
$W=h v=\frac{h c}{\lambda}$
Where $v$ is frequency, $c$ is speed of light, $\lambda$ is wavelength.
Given, $W=2\, e V $
$=2 \times 1.6 \times 10^{-19}\, J$
$h=6.6 \times 10^{-34} \,J-s, $
$c=3 \times 10^{8} \,m / s$
$\Rightarrow \lambda=\frac{h c}{W} $
$=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{2 \times 1.6 \times 10^{-19}}$
$=620 \times 10^{-9} \,m =620\, nm$
Note: If wavelength of incident light greater than $620 \,nm$ no photoelectrons are emitted.