Question

The photoelectric threshold wavelength for potassium having work function of 2 eV, is
( Take h = $6.6\times 10^{-34}J-s,1eV=1.6\times 10^{-19}J,c=3\times 10^{8}m{s}^{-1})$

• A. 1860 nm
• B. 618.7 nm
• C. 1240 nm
• D. 317.0 nm

Answer 1

Answer: (B)

We know that, work function
$\varphi =h{v}_0$
$\Rightarrow \varphi =\frac{hc}{{\lambda }_0}$ $[\because v_0=\frac{c}{{\lambda }_0}]$
$\Rightarrow {\lambda }_0=\frac{hc}{\varphi }$
$\therefore =\frac{6.6\times 10^{-34}\times 3\times 10^8}{2\times 1.6\times 10^{-19}}$
= $\frac{6.6\times 3\times 10^{-7}}{2\times 1.6}$
= $6.187\times 10^{-7}$
= 618.7 nm