A hydrogen atom changes its state from n=3 to | vedclass

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(B) The energy difference for the transition $n=3$ to $n=2$ is $\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}ight) = 1.9 	ext{ eV}$.The emitt

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$7$

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(B) The energy difference for the transition $n=3$ to $n=2$ is $\Delta E = 13.6 \left(\frac{1}{2^2} - \frac{1}{3^2}ight) = 1.9 	ext{ eV}$.The emitted photon has wavelength $\lambda = \frac{hc}{\Delta E}$.Due to conservation of momentum,the atom recoils with velocity $v$,so $mv = \frac{h}{\lambda'}$,where $\lambda'$ is the wavelength of the emitted photon considering recoil.The energy balance is $\Delta E = \frac{1}{2}mv^2 + \frac{hc}{\lambda'}$.Substituting $v = \frac{h}{m\lambda'}$,we get $\Delta E = \frac{h^2}{2m\lambda'^2} + \frac{hc}{\lambda'}$.Solving for $\lambda'$,we find $\lambda' \approx \lambda \left(1 + \frac{\Delta E}{2mc^2}ight)$.The fractional change in wavelength is $\frac{\Delta \lambda}{\lambda} = \frac{\lambda' - \lambda}{\lambda} = \frac{\Delta E}{2mc^2}$.Substituting the values: $\Delta E = 1.9 	imes 1.6 	imes 10^{-19} 	ext{ J}$,$m = 1.67 	imes 10^{-27} 	ext{ kg}$,$c = 3 	imes 10^8 	ext{ m/s}$.$\frac{\Delta \lambda}{\lambda} = \frac{1.9 	imes 1.6 	imes 10^{-19}}{2 	imes 1.67 	imes 10^{-27} 	imes (3 	imes 10^8)^2} \approx \frac{3.04 	imes 10^{-19}}{3.006 	imes 10^{-10}} \approx 10^{-9}$.Since the question asks for percentage change,$\% 	ext{ change} = \frac{\Delta \lambda}{\lambda} 	imes 100 \approx 10^{-9} 	imes 10^2 = 10^{-7}$.Thus,$n = 7$.

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