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Q.
Calcium plate has maximum possible radiation of wavelength $ \lambda $ of $400 \,nm$ to eject electrons. Its work function is
ManipalManipal 2010Dual Nature of Radiation and Matter
Solution:
The minimum energy required for the emission of photoelectron from a metal is called the work function $W$ of that metal. If $v_{0}$ is threshold frequency, then
$ W =h v_{0} $
but frequency $\left(v_{0}\right) =\frac{\text { velocity }(c)}{\text { wavelength }(\lambda)} $
$\therefore W =\frac{h c}{\lambda}$
where, $h$ is Planck's constant and $c$ the speed of light.
Given, $ h=6.6 \times 10^{-34} Js , c=3 \times 10^{8} m / s ,$
$ \lambda=400 \,nm =400 \times 10^{-9} m $
$ W=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{400 \times 10^{-9}} J$
$ W =4.95 \times 10^{-19} J $
$W =\frac{4.95 \times 10^{-19}}{1.6 \times 10^{-19}}=3.1 \,eV$