JEE Advance - Physics (2011 - Paper 1 Offline - No. 2)
Explanation
For resonance
$$\omega = {\omega _p} = \sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}} $$ [From previous question]
$$\upsilon = {\omega \over {2\pi }} = {1 \over {2\pi }}\sqrt {{{N{e^2}} \over {m{\varepsilon _0}}}} $$
$$\lambda = {c \over \upsilon } = 2\pi c\sqrt {{{m{\varepsilon _0}} \over {N{e^2}}}} $$
Substituting the given values, we get
$$\lambda = 2 \times 3.14 \times 3 \times {10^8} \times \sqrt {{{{{10}^{ - 30}} \times {{10}^{ - 11}}} \over {4 \times {{10}^{27}} \times {{(1.6 \times {{10}^{ - 19}})}^2}}}} $$
$$ = {{2 \times 3.14 \times 3 \times {{10}^8}} \over {1.6 \times {{10}^{ - 19}}}}\sqrt {{{{{10}^{ - 41}}} \over {4 \times {{10}^{27}}}}} = {{9.42} \over {1.6}} \times {10^{27}} \times {10^{ - 34}}$$ m
$$ \approx 6.00 \times {10^{ - 7}}$$ m = 600 nm
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