A) 0.9 V
B) 0.5 V
C) 0.4 V
D) 0.1 V
Correct Answer: A
Solution :
As we know in Young's double slit experiment fringe width = separation between two consecutive fringe or dark fringes \[=\beta =\frac{\lambda D}{d}\] Here \[\beta =2y\] Þ \[2y=\frac{\lambda D}{d}\] Þ \[\lambda =\frac{2yd}{D}\] Þ \[\lambda =\frac{2\times 1\times {{10}^{-3}}\times 0.24\times {{10}^{-3}}}{1.2}=4\times {{10}^{-7}}m=4000\,{\AA}\] Energy of light incident on photo plate \[E\,(eV)=\frac{12375}{4000}=3.1\,eV\] According to Eienstein photoelectric equation E = W0 + eV0 Þ \[{{V}_{0}}=\frac{(E-{{W}_{0}})}{e}=\frac{(3+2.2)}{e}eV\,\approx 0.9\,V\]
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