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Q. What is the work function of the metal if the light of wavelength $4000\, \mathring A$ generates photoelectrons of velocity $6 \times 10^5 \; ms^{-1}$ form it ?
(Mass of electron = $9 \times 10^{-31} \; kg$
Velocity of light = $3 \times 10^8 \; ms^{-1}$
Planck's constant = $6.626 \times 10^{-34} \; Js$
Charge of electron = $1.6 \times 10^{-19} \; JeV^{-1}$)

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Solution:

$hv =\phi + hv^{\circ} $
$ \frac{1}{2}mv^{2} = hc \left(\frac{1}{\lambda} - \frac{1}{\lambda_{0}}\right) $
$ hv =\phi + \frac{1}{2} mv^{2}$
$ \phi = \frac{6.626 \times10^{-34}\times3 \times10^{8}}{4000\times10^{-10}} - \frac{1}{2} \times9 \times10^{-31} \times\left(6\times10^{5}\right)^{2} $
$ \phi = 3.35 \times10^{-19 } J \Rightarrow \phi \simeq 2.1 eV $