Question

The work function of a metal is 3.0eV. It is illuminated by a light of wave length $3\times {10}^{-7}m$. then
i) Threshold frequency
ii) maximum energy of photoelectrons,
iii) the stopping potential.

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

• A. $0.72\times {10}^{15}Hz,1.81\times {10}^{-19}J,1.13V$
• B. $0.42\times {10}^{15}Hz,1.96\times {10}^{-19}J,2.13V$
• C. $0.89\times {10}^{15}Hz,3.92\times {10}^{-19}J,6.14V$
• D. $1.67\times {10}^{15}Hz,2.3\times {10}^{-19}J,3.1V$

Answer 1

The correct option is A $0.72\times {10}^{15}Hz,1.81\times {10}^{-19}J,1.13V$

Threshold frequency $=\frac{\varphi }{h}$

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

$=0.72\times {10}^{15}HZ.$

maximum energy $=\frac{hc}{\lambda }-\varphi$

$=\frac{1240}{300}-3(hc=1240ev-nm)$

$=1.13ev.$
$=1.13\times 1.6\times {10}^{-19}$
$=1.81\times {10}^{-19}J.$
So, stopping potential
$=maximumenergy/e$
$=1.13ev/e$
$=1.13v$
So, the answer is (A).