A) B alone
B) A alone
C) Neither A nor B
D) Both A & B
Correct Answer: B
Solution :
\[{{w}_{A}}=\frac{6.6\times {{10}^{-34}}\,\times 1.8\times {{10}^{14}}}{1.6\times {{10}^{-19}}}\,=\,0.74\,\,eV\]\[{{w}_{B}}=\frac{6.6\times {{10}^{-34}}\,\times 2.2\times {{10}^{14}}}{1.6\times {{10}^{-19}}}\,=\,0.91\,\,eV\] \[\therefore \] photoelectrons will emit from A aloneYou need to login to perform this action.
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