The radiation corresponding to 3 rightarrow 2 | vedclass

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(A) The energy of the photon emitted during the $3 \rightarrow 2$ transition in a hydrogen atom is given by:$E = 13.6 \left( \frac{1}{2^2} - \frac{1}{

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$1.1$

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(A) The energy of the photon emitted during the $3 ightarrow 2$ transition in a hydrogen atom is given by:$E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 	imes \frac{5}{36} \approx 1.889 \ eV$.The radius of the circular path of an electron in a magnetic field is $r = \frac{mv}{qB} = \frac{\sqrt{2mK}}{eB}$,where $K$ is the kinetic energy of the photoelectron.$K = \frac{r^2 e^2 B^2}{2m}$.Given $r = 10.0 \ mm = 10^{-2} \ m$,$B = 3 	imes 10^{-4} \ T$,$e = 1.6 	imes 10^{-19} \ C$,and $m = 9.1 	imes 10^{-31} \ kg$:$K = \frac{(10^{-2})^2 	imes (1.6 	imes 10^{-19}) 	imes (3 	imes 10^{-4})^2}{2 	imes 9.1 	imes 10^{-31}} \approx 0.79 \ eV \approx 0.8 \ eV$.Using Einstein's photoelectric equation: $E = \Phi + K$,where $\Phi$ is the work function.$\Phi = E - K = 1.889 \ eV - 0.8 \ eV = 1.089 \ eV \approx 1.1 \ eV$.

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