$\begin{array}{1 1}(a)\;1.2\times 10^{-7}m&(b)\;1.2\times 10^{-9}m\\(c)\;2.3\times 10^{-9}m&(d)\;1\times 10^{-3}m\end{array}$

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+1 vote

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KE =Quantum energy-Threshold energy

$\Rightarrow \large\frac{6.626\times 10^{-34}\times 3\times 10^8}{3000\times 10^{-10}}-\frac{6.626\times 10^{-34}\times 3\times 10^8}{4000\times 10^{-10}}$

$\Rightarrow 6.626\times 10^{-19}-4.9695\times 10^{-19}$

$\Rightarrow 1.6565\times 10^{-19}$J

$\large\frac{1}{2}$$mv^2=1.6565\times 10^{-19}$

$m^2v^2=2\times 1.6565\times 10^{-19}\times 9.1\times 10^{-31}$

$mv=5.49\times 10^{-25}$

$\lambda=\large\frac{h}{mv}$

$\Rightarrow \large\frac{6.626\times 10^{-34}}{5.49\times 10^{-25}}$

$\Rightarrow 1.2\times 10^{-9}m$

Hence (b) is the correct answer.

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