A) \[1:4\]
B) \[1:1\]
C) \[1:5\]
D) \[1:2\]
Correct Answer: D
Solution :
: According to Einsteins photoelectric equation, the maximum kinetic energy of emitted photoelectrons is \[{{K}_{\max }}=h\upsilon -{{\phi }_{0}}\] where \[h\upsilon \] is the energy of incident photon and \[{{\phi }_{0}}\]is the work function. But \[{{K}_{\max }}=\frac{1}{2}mv_{\max }^{2}\] \[\therefore \] \[\frac{1}{2}mv_{\max }^{2}=h\upsilon -{{\phi }_{0}}\] As per question \[\frac{1}{2}mv_{{{\max }_{1}}}^{2}=1eV-0.5eV=0.5eV\]?.(i) and \[\frac{1}{2}mv_{{{\max }_{2}}}^{2}=2.5eV-0.5eV=2eV\] ?.(ii) Dividing eqn. (i) by eqn. (ii), we get \[\frac{v_{{{\max }_{1}}}^{2}}{v_{{{\max }_{2}}}^{2}}=\frac{0.5eV}{2eV}=\frac{1}{4}\] \[\frac{{{v}_{{{\max }_{1}}}}}{{{v}_{{{\max }_{2}}}}}=\sqrt{\frac{1}{4}}=\frac{1}{2}\]
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