Question

If the work function for a certain metal is $3.2\times 10^{- 19 \, }J$ and it is illuminated with light of frequency $\nu=8\times 10^{14 \, } \, Hz$ , the maximum kinetic energy of the photoelectron would be

• A. $2.1\times 10^{- 19 \, }J$
• B. $3.2\times 10^{- 19 \, }J$
• C. $5.3\times 10^{- 19 \, }J$
• D. $8.5\times 10^{- 19 \, }J$

Answer 1

Answer: (A)

Maximum KE $=hv-\phi_{0}$
$=6.63\times 10^{- 34}\times 8\times 10^{14}-3.2\times 10^{- 19}$
$=2.1\times 10^{- 19 \, }J$