A) \[{{10}^{14}}\] and 10 eV
B) \[{{10}^{12}}\] and 5 Ev
C) \[{{10}^{11}}\] and 5 eV
D) \[{{10}^{10}}\]and 5 eV
Correct Answer: D
Solution :
Energy incident on plate per second \[=\,\,IA\] \[~=16\times 10-3\times 1\times {{10}^{-}}^{4}\] \[=\,\,16\times {{10}^{-}}^{7}watt\] \[K{{E}_{max}}=hf-\phi \] \[=\text{ }10-5=5\,eV\] \[\frac{\overset{\bullet }{\mathop{N}}\,\,hc}{\lambda }=1.6\times {{10}^{-\,7}}\] \[\overset{\bullet }{\mathop{N}}\,\times 10\times 1.6\times {{10}^{-19}}\,=1.6\times {{10}^{-7}}\] \[\overset{\bullet }{\mathop{N}}\,\times \text{1}{{\text{0}}^{-18}}\,={{10}^{-7}}\] \[\overset{\bullet }{\mathop{N}}\,={{10}^{11}}\] \[10%\] of incident photon emit electron \[\therefore \] No of emitted electrons per second \[=\,\,\overset{\bullet }{\mathop{N}}\,\,\times \,\frac{10}{100}\,\,=\,\,{{10}^{10}}\]You need to login to perform this action.
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