Question

Ultraviolet radiation of 6.2 eV falls on an aluminium surface (work function 4.2 eV). The kinetic energy (in joule) of the fastest electron emitted is :

• A. $3.2\times {10}^{-21}$
• B. $1.6\times {10}^{-17}$
• C. $3.2\times {10}^{-19}$
• D. $3.2\times {10}^{-15}$

Answer 1

The correct option is C $3.2\times {10}^{-19}$

$Energyofradiation,h\nu =6.2eV=6.2\times 1.6\times {10}^{-19}JWorkfunctionW=4.2eV=4.2\times 1.6\times {10}^{-19}JKE=h\nu -W=6.2\times 1.6\times {10}^{-19}-4.2\times 1.6\times {10}^{-19}=2\times 1.6\times {10}^{-19}=3.2{\times 10}^{-19}J$