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Q. The surface of a metal is illuminated with a light of wavelength $400 \, nm$ . The maximum kinetic energy of the ejected photoelectrons was found to be $1.68 \, eV$ . The work function of the metal is $\left(h c = 1240 \, e V \, n m\right)$

NTA AbhyasNTA Abhyas 2020

Solution:

$\because \, \, \frac{h c}{\lambda }=\frac{1}{2}mv^{2}+\phi$
$\Rightarrow \, \, \phi=\frac{h c}{\lambda }-\frac{1}{2}mv^{2}$
$\phi=\frac{1240}{400}-1.68=1.41 \, eV$