A) \[\text{3}.\text{2}\times \text{I}{{\text{0}}^{\text{-21}}}\text{ J}\]
B) \[\text{3}.\text{2}\times \text{1}{{0}^{-19}}\text{ J}\]
C) \[\text{3}.\text{2}\times \text{1}{{0}^{-\text{15}}}\text{ J}\]
D) \[\text{3}.\text{2}\times \text{l}{{\text{O}}^{-17}}\text{ J}\]
Correct Answer: B
Solution :
From Einsteins relation, the kinetic energy (KE) of emitted electron is given by \[KE=hv-{{W}_{0}}\] where, \[{{W}_{0}}\] is work function of metal, v the frequency, h the Plancks constant. Given, \[hv=6.2\,eV\], \[{{W}_{0}}=4.2\,eV\] \[\therefore \] \[KE=6.2-4.2=2\,eV\] Also, \[1\,eV=1.6\times {{10}^{-19}}\,J\] \[\therefore \] \[KE=2\times 1.6\times {{10}^{-19}}J\] \[=3.2\times {{10}^{-19}}J\]
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