A) B alone
B) A alone
C) neither A nor B
D) both A and B
Correct Answer: B
Solution :
Threshold energy of A is \[{{E}_{A}}=h{{v}_{A}}\] \[=6.6\times {{10}^{-34}}\times 1.8\times {{10}^{14}}\] \[=11.88\times {{10}^{-20}}J\] \[=\frac{11.88\times {{10}^{-20}}}{1.6\times {{10}^{-19}}}eV\] \[=0.74eV\] Similarly, \[{{E}_{B}}=0.91eV\] Since, the incident photons have energy greater than \[{{E}_{A}}\]but less than\[{{E}_{B}}.\] So, photoelectrons will be emitted from metal A only.You need to login to perform this action.
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