Question

Ultraviolet radiation of 6.2 eV falls on an aluminium foil surface. Work function is 4.2 eV. The K.E. of the fastest electron emitted approximately :

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

Answer 1

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

K.E. = Energy of radiation(hv) - Work function $\left(h{v}_0\right)$
$=6.2eV-4.2eV$
$=2eV=2\times 1.6\times {10}^{-19}J$
$=3.2\times {10}^{-19}J$