A) 11.5eV
B) 11.5keV
C) 125eV
D) 1250eV
Correct Answer: C
Solution :
Given that Potential difference V = 125 V Then, \[\lambda =\frac{12375}{V}\overset{o}{\mathop{A}}\,\] \[\lambda =\frac{12375}{125}\overset{o}{\mathop{A}}\,\] \[\therefore \] Energy of electron in eV \[E=\frac{hc}{\lambda }\] \[E=\frac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}\times 125\times {{10}^{-10}}}{12375}\] = 125 eV Then the energy of photon = 125 eVYou need to login to perform this action.
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