$(a)\;813.4 \overset{\circ}{A}\qquad(b)\;672.4 \overset{\circ}{A}\qquad(c)\;912.24\overset{\circ}{A} \qquad d)\;734 \overset{\circ}{A}\qquad$

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The photon capable of removing electron from I Bohr's orbit must posses energy

= 13.6 eV

= $13.6\times1.602\times10^{-19}$J

= $21.787\times10^{-19}$ J

E = $\large\frac{h\times c}{\lambda}$

$21.787\times10^{-19}$ = $\large\frac{6.625\times10^{-34}\times3.0\times10^8}{\lambda}$

$\lambda = 912.24\times10^{-10}$ m

= 912.24 $\overset{\circ}{A}$

Hence the answer is (c)

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