A difference of 2.3; eV separates two energy | vedclass

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Question

(B) The energy difference between the two levels is given as $E = 2.3 \; eV$.To convert this energy into Joules,we multiply by the charge of an electr

Vedclass · Accepted Answer

$5.6 	imes 10^{14}\; Hz$.

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(B) The energy difference between the two levels is given as $E = 2.3 \; eV$.To convert this energy into Joules,we multiply by the charge of an electron $(1.6 	imes 10^{-19} \; C)$:$E = 2.3 	imes 1.6 	imes 10^{-19} \; J = 3.68 	imes 10^{-19} \; J$.According to the Bohr model,the frequency $
u$ of the emitted radiation is related to the energy difference by the equation $E = h
u$,where $h$ is Planck's constant $(6.626 	imes 10^{-34} \; Js)$.Rearranging for frequency: $
u = \frac{E}{h}$.Substituting the values: $
u = \frac{3.68 	imes 10^{-19}}{6.626 	imes 10^{-34}} \approx 5.55 	imes 10^{14} \; Hz$.Rounding to two significant figures,we get $
u \approx 5.6 	imes 10^{14} \; Hz$.

$A$ difference of $2.3\; eV$ separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?

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