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VITEEE 2009 Paper

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Question : 33 of 120

Marks: +1, -0

The work function of a certain metal is

3.31\times10^{-19}\,\mathrm{J}

Then, the maximum kinetic energy of photoelectrons emitted by incident radiation of wavelength 5000 Å is (Given,

h=6.62\times10^{-34}\,\mathrm{J\,s},c=3\times10^{8}\,\mathrm{m\,s^{-1}},e=1.6\times10^{-19}\,\mathrm{C}

)

2.48 eV
o.41 eV
2.07 eV
0.82 eV

Solution:

Work function

W_{0}=3.31\times10^{-19}\,\mathrm{J}

Wavelength of incident radiation

\lambda=5000\times10^{-10}

m

According to Einstein's photoelectric equation

E=W_{0}+KE

⇒

\frac{hc}{\lambda}=3.31\times10^{-19}+KE

⇒

KE=-3.31\times10^{-19}+\frac{6.62\times10^{-34}\times3\times10^{8}}{5000\times10^{-10}}

=-3.31\times10^{-19}+\frac{6.62\times3}{5}\times10^{-19}

=(-3.31\times1.324\times3)\times10^{-19}

=(3.972-3.31)\times10^{-19}

=0.662\times10^{-19}\,\mathrm{J}

or

E=\frac{0.662\times10^{-19}}{1.6\times10^{-19}}=0.41\,\mathrm{eV}

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