A) 151V
B) 15.3V
C) 12.3V
D) 13.6V
Correct Answer: A
Solution :
Stopping potential (Vo) is the minimum negative (retarding) potential given to the anode for which the photoelectric current becomes zero. Photoelectric current becomes zero when the stopping potential equals the maximum kinetic energy \[({{K}_{\max }})\]the photoelectrons. That is, \[{{K}_{\max }}=e{{V}_{0}}\] ... (i) Now, de-Broglie wavelength of electron is given by \[{{\lambda }_{e}}=\frac{h}{\sqrt{2m{{K}_{\max }}}}\] ... (ii) Equating Eqs. (i) and (ii), we arrive at \[{{\lambda }_{e}}=\frac{h}{\sqrt{2me{{V}_{0}}}}\] \[\Rightarrow \] \[{{V}_{0}}=\frac{{{h}^{2}}}{2me\lambda _{e}^{2}}\] ?. (ii) Given,\[{{\lambda }_{e}}=1\,{{A}^{o}}=1\times {{10}^{-10}}m\], \[h=6.6\times {{10}^{-34}}J-s\]. \[m=9.1\times {{10}^{-31}}kg\], \[e=1.6\times {{10}^{-19}}C\] \[\therefore \] \[{{V}_{0}}=\frac{{{(6.6\times {{10}^{-34}})}^{2}}}{2\times (9.1\times {{10}^{-31}})\times (1.6\times {{10}^{-19}})}\] \[\times {{(1\times {{10}^{-10}})}^{2}}\] = 151 volt. Note: For a given frequency of the incident radiation, the stopping potential is independent of its intensity.
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