A) metal A
B) metal B
C) both A and B
D) neither metal A nor metal B
Correct Answer: B
Solution :
The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal. From Plancks law energy (E) of incident light is \[E=\frac{hc}{\lambda }\] where\[\lambda \]is wavelength. Given,\[\lambda =3500{\AA}=3500\times {{10}^{-10}}m,\] \[h=6.64\times {{10}^{-34}}J-s,\] \[c=3\times {{10}^{8}}m/s\] \[E=\frac{6.64\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3500\times {{10}^{-10}}}\] \[=5.69\times {{10}^{-19}}\,J\] Also, \[1\,eV=1.6\times {{10}^{-19}}J\] \[\therefore \] \[E=\frac{5.69\times {{10}^{-19}}}{1.6\times {{10}^{-19}}}eV=3.55\,eV\] Hence, only B will emit photoelectrons as incident energy is greater than work function.
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