$(a)\;68.73nm\qquad(b)\;50.71nm\qquad(c)\;47nm\qquad(d)\;52.3nm$

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This is a case of photoelectric effect, work function being the ionization energy.

$\therefore$ Total Energy = KE + work function

$\large\frac{hc}{\lambda} = \large\frac{1}{2}mv^2+w$

$\large\frac{6.626\times10^{-34}\times3\times10^8}{\lambda} = \large\frac{1}{2}\times 9.1\times10^{-31}\times(1.03\times10^6)^2+3.44\times10^{-18}$

$\large\frac{2\times10^{-25}}{\lambda} = 3.92\times10^{-18}$

$\lambda = \large\frac{2\times10^{-25}}{3.92\times10^{-18}}$

=50.71nm

Hence the answer is (b)

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