Explanation
Given that,
Band gap of $${{D}_{1}}=2.5\,eV$$
Band gap of $${{D}_{2}}=2\,eV$$
Band gap of $${{D}_{3}}=3\,eV$$
Wave length $$\lambda =6000\overset{\circ }{\mathop{A}}\,$$
Now, the wave length for $$2.5\ eV$$
$$ {{\lambda }_{1}}=\dfrac{hc}{E} $$
$$ {{\lambda }_{1}}=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2.5\times 1.6\times {{10}^{-19}}} $$
$$ {{\lambda }_{1}}=4.95\times {{10}^{-7}}\,m $$
Now, for $$2\ eV$$
$$ {{\lambda }_{2}}=\dfrac{hc}{E} $$
$$ {{\lambda }_{2}}=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{2\times 1.6\times {{10}^{-19}}} $$
$$ {{\lambda }_{2}}=6.2\times {{10}^{-7}}\,m $$
Now, for $$3\ eV$$
$$ {{\lambda }_{3}}=\dfrac{hc}{E} $$
$$ {{\lambda }_{3}}=\dfrac{6.6\times {{10}^{-34}}\times 3\times {{10}^{8}}}{3\times 1.6\times {{10}^{-19}}} $$
$$ {{\lambda }_{3}}=4.13\times {{10}^{-7}}\,m $$
For detection of optical signal the wave length of incident energy radiation must be greater.
So, only $${{D}_{2}}$$can detect the radiation
The radius has a ratio of $$2:1$$, therefore the cross-sectional area is of ratio $$4:1.$$
The resistance per unit length is therefore $$ 1:4.$$
Because we are told the mass is the same, wire $$B$$, the smaller section wire will have to be $$4$$ times as long to have the same mass, neglecting the effects of any insulation on the mass.
So the final resistance of $$ B$$ will be $$16$$ times the resistance of $$A$$ or $$544$$ Ohms.
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