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(A) The electric field $E$ at a distance $r$ from the axis of a charged cylinder of length $L$ and charge $Q$ is given by $E = \frac{\lambda}{2 \pi \v

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Infrared

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(A) The electric field $E$ at a distance $r$ from the axis of a charged cylinder of length $L$ and charge $Q$ is given by $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$,where $\lambda = Q/L$ is the linear charge density.The potential $V(r)$ is $V(r) = -\int E \, dr = -\frac{\lambda}{2 \pi \varepsilon_0} \ln(r) + C$.The potential energy of an electron in this field is $U(r) = -eV(r) = \frac{e \lambda}{2 \pi \varepsilon_0} \ln(r) = 2k \lambda e \ln(r)$.Given $Q = 10e = 1.6 	imes 10^{-18} \,C$ and $L = 10^{-6} \,m$,$\lambda = 1.6 	imes 10^{-12} \,C/m$.The energy levels in this logarithmic potential are analogous to the Bohr model where the transition energy $\Delta E = E_2 - E_1 = 2k \lambda e \ln(r_2/r_1)$. For the first transition,$\Delta E = 2k \lambda e \ln(2)$.Substituting the values: $\Delta E = 2 	imes (9 	imes 10^9) 	imes (1.6 	imes 10^{-12}) 	imes (1.6 	imes 10^{-19}) 	imes 0.693 \approx 3.19 	imes 10^{-21} \,J$.Frequency $f = \frac{\Delta E}{h} = \frac{3.19 	imes 10^{-21}}{6.63 	imes 10^{-34}} \approx 4.8 	imes 10^{12} \,Hz$.This frequency corresponds to the Infrared region of the electromagnetic spectrum.

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