Question

The minimum energy required to overcome the attractive force between an electron and the surface of $Ag$ metal is $7.52\times {10}^{-19}J$ . What will be the maximum kinetic energy of electrons ejected out from $Ag$ which is being exposed to UV light of $\lambda =360\stackrel{\circ }{A}$.

• A. $47.68\times {10}^{19}J$
• B. $47.68\times {10}^{-19}J$
• C. $47.68\times {10}^{-18}J$
• D. $47.68\times {10}^{18}J$

Answer 1

The correct option is B $47.68\times {10}^{-19}J$

We know, from photoelectric effect,
Kinetic energy (K.E.) of photoelectron = Energy of photon - Work function
$K.E=h\nu -\varphi$ = $\frac{hc}{\lambda }-\varphi$
Given: Work function = $7.52\times {10}^{-19}J$
where h = Planck's constant
$\nu$ = incident frequency
Put up the values,
K.E = $\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{360\times {10}^{-10}}-7.52\times {10}^{-19}$
On solving,
K.E = $47.68\times {10}^{-19}$ J