Correct Answer - Option 1 : 7692 Å
From Einstein’s photoelectric equation,
\(\begin{array}{l}
h\nu = h{\nu _0} + \frac{1}{2}mv_{max}^2\;,\;\;{\nu _0} = threshold\;frequency\\
hc\left( {\frac{1}{\lambda } - \frac{1}{{{\lambda _0}}}} \right) = \frac{1}{2}mv_{max}^2\\
\frac{1}{{{\lambda _0}}} = \frac{1}{\lambda } - \frac{1}{2}\frac{{9.11 \times {{10}^{ - 31}} \times {{\left( {4 \times {{10}^5}} \right)}^2}}}{{6.626 \times {{10}^{ - 34}} \times 3 \times {{10}^8}}}\\
= \frac{1}{{6000 \times {{10}^{ - 10}}}} - 0.0366 \times {10^7} = 0.1666 \times {10^7} - 0.0366 \times {10^7}\\
\Rightarrow {\lambda _0} = 7.692 \times {10^{ - 7}}m = 7692 \times {10^{ - 10}}m = 7692\;\mathop A\limits^ \circ
\end{array}\)