Question

The work function of a metal is 2.5 eV. The maximum kinetic energy of the photoelectrons emitted if a radiation of wavelength 3000$A^o$ falls on it is :

($h=6.63\times {10}^{-34}$ Js and $c=3\times {10}^8$ m/s)

• A. $1.12\times {10}^{-19}J$
• B. $4.8\times {10}^{-19}J$
• C. $3.2\times {10}^{-19}J$
• D. $2.63\times {10}^{-19}J$

Answer 1

The correct option is D $2.63\times {10}^{-19}J$

Maximum kinetic energy, ${K.E.}_{max}=\frac{hc}{\lambda }-\varphi$

$=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{3000\times {10}^{-10}}-2.5\times 1.6\times {10}^{-19}$

$=(6.63\times {10}^{-19}-4\times {10}^{-19})J$

$=2.63\times {10}^{-19}J$.

So, the answer is option (D).