JEE Advance - Physics (2011 - Paper 1 Offline - No. 2)

Estimate the wavelength at which plasma reflection will occur for a metal having the density of electrons N $$ \approx $$ 4 $$ \times $$ 1027 m-3. Taking $${{\varepsilon _0}}$$ = 10- 11 and m $$ \approx $$ 10- 30, where these quantities are in proper SI units.
800 nm
600 nm
300 nm
200 nm

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

Comments (0)

Advertisement