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Q. If wavelength $ \lambda =5400\,\mathring{A}$ is threshold value for a certain metal, then its work function would be:

Rajasthan PMTRajasthan PMT 2006Dual Nature of Radiation and Matter

Solution:

$ E=\frac{h c}{\lambda} J =\frac{. h c}{\lambda \times 1.6 \times 10^{-19}} eV$
$\Rightarrow E=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{5400 \times 10^{-10} \times 1.6 \times 10^{-19}} eV $
$ \Rightarrow E=2.3 \,eV$