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If a light of 6000 Å falls on a metal surface and emits photoelectrons with velocity of 4 x 105 m/sec. What is the threshold wavelength? (Plank’s constant = 6.626 x 10-34 J.sec )
1. 7692 Å
2. 5000 Å
3. 9692 Å
4. None

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Correct Answer - Option 1 : 7692 Å

From Einstein’s photoelectric equation,

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

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