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

Answer: (B)

Using the relation $ KE=hv-{{\omega }_{0}} $ $ KE=6.2-4.2\,=2\,eV $ $ =2\times 1.6\times {{10}^{-19}}=3.2\times {{10}^{-19}}\,J $