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(A) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda

Vedclass · Accepted Answer

$4 \lambda$

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(A) According to Einstein's photoelectric equation: $eV = \frac{hc}{\lambda} - \frac{hc}{\lambda_{0}} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$.For the first case: $3 eV_{s} = hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)$  --- $(i)$For the second case: $eV_{s} = hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)$  --- (ii)Dividing equation $(i)$ by equation (ii):$\frac{3 eV_{s}}{eV_{s}} = \frac{hc \left( \frac{1}{\lambda} - \frac{1}{\lambda_{0}} \right)}{hc \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right)}$$3 = \frac{\frac{1}{\lambda} - \frac{1}{\lambda_{0}}}{\frac{1}{2 \lambda} - \frac{1}{\lambda_{0}}}$$3 \left( \frac{1}{2 \lambda} - \frac{1}{\lambda_{0}} \right) = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$$\frac{3}{2 \lambda} - \frac{3}{\lambda_{0}} = \frac{1}{\lambda} - \frac{1}{\lambda_{0}}$$\frac{3}{2 \lambda} - \frac{1}{\lambda} = \frac{3}{\lambda_{0}} - \frac{1}{\lambda_{0}}$$\frac{1}{2 \lambda} = \frac{2}{\lambda_{0}}$$\lambda_{0} = 4 \lambda$.

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