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Question

(A) The energy of a 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}{3^

Vedclass · Accepted Answer

$0.82$

Vedclass · Answer

(A) The energy of a photon emitted during the $3 ightarrow 2$ transition in a hydrogen atom is given by $E = 13.6 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 13.6 \left( \frac{1}{4} - \frac{1}{9} ight) = 13.6 	imes \frac{5}{36} \approx 1.89 \, eV$.The kinetic energy $(K.E.)$ of the photoelectrons is related to the radius $r$ of their circular path in a magnetic field $B$ by $r = \frac{mv}{qB}$,where $p = mv = qBr$.Thus,$K.E. = \frac{p^2}{2m} = \frac{(qBr)^2}{2m}$.Given $q = 1.6 	imes 10^{-19} \, C$,$B = 5 	imes 10^{-4} \, T$,and $r = 7 	imes 10^{-3} \, m$:$p = (1.6 	imes 10^{-19}) 	imes (7 	imes 10^{-3}) 	imes (5 	imes 10^{-4}) = 5.6 	imes 10^{-25} \, kg \cdot m/s$.$K.E. = \frac{(5.6 	imes 10^{-25})^2}{2 	imes 9.1 	imes 10^{-31}} = \frac{31.36 	imes 10^{-50}}{18.2 	imes 10^{-31}} \approx 1.723 	imes 10^{-19} \, J$.Converting to $eV$: $K.E. = \frac{1.723 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 1.077 \, eV$.Using Einstein's photoelectric equation: $\Phi = E_{photon} - K.E._{max} = 1.89 \, eV - 1.077 \, eV = 0.813 \, eV \approx 0.82 \, eV$.

Column-$I$	Column-$II$
$A$. Hallwachs $\&$ Lenard	$P$. Transformers
$B$. Franck-Hertz	$Q$. Microwave
$C$. Klystron valve	$R$. Quantization of energy levels
$D$. Nicola Tesla	$S$. Photoelectric effect

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In the Thomson spectrograph experiment,four positive ions $P, Q, R,$ and $S$ are situated on the $Y-X$ curves as shown in the figure.

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$A$. Hallwachs $\&$ Lenard $P$. Transformers

$B$. Franck-Hertz $Q$. Microwave

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