A) Lithium
B) Copper
C) Both
D) None
Correct Answer: A
Solution :
Key Idea: The approximate range of visible spectrum is 400 to 700 nm. The minimum energy required for the emission of photoelectrons from a metal is called the work function of that metal. \[W=\frac{hc}{\lambda }\] \[\Rightarrow \] \[\lambda =\frac{hc}{W}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.3\times 1.6\times {{10}^{-19}}}\] \[=5.38\times {{10}^{-7}}m\] = 538 nm (for lithium) For copper \[\lambda =\frac{hc}{W}=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{4\times 1.6\times {{10}^{-19}}}=309\,nm\] Since, wavelength range of visible spectrum is 400 nm to 700 nm, have only lithium lies in this range. Hence, lithium is useful.
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