Question

The work function of a metal is $1.6\times {10}^{-19}$J. When the metal surface is illuminated by the light of wavelength 6400 $A^o$, then the maximum kinetic energy of emitted photoelectrons will be ($h=6.4\times {10}^{-34}Js$)

• A. $14\times {10}^{-19}J$
• B. $2.8\times {10}^{-19}J$
• C. $1.4\times {10}^{-19}J$
• D. $1.4\times {10}^{-19}eV$

Answer 1

The correct option is C $1.4\times {10}^{-19}J$

$K.E{.}_{max}=\frac{hc}{\lambda }-\varphi$

$=\frac{6.4\times {10}^{-34}\times 3\times {10}^8}{6.4\times {10}^{-7}}-1.6\times {10}^{-19}$

$=3\times {10}^{-19}-1.6\times {10}^{-19}$
$=1.4\times {10}^{-19}J.$
So, the answer is option (C).