A difference of 5.4 eV separates two energy | vedclass

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Question

(B) The energy difference between the two levels is given by $\Delta E = 5.4 \ eV$.Converting this energy into Joules: $\Delta E = 5.4 	imes 1.6 	im

Vedclass · Accepted Answer

$1.304 	imes 10^{15} \ Hz$

Vedclass · Answer

(B) The energy difference between the two levels is given by $\Delta E = 5.4 \ eV$.Converting this energy into Joules: $\Delta E = 5.4 	imes 1.6 	imes 10^{-19} \ J = 8.64 	imes 10^{-19} \ J$.The energy of the emitted photon is related to the frequency $
u$ by the equation $\Delta E = h
u$.Therefore,the frequency $
u = \frac{\Delta E}{h}$.Substituting the values: $
u = \frac{8.64 	imes 10^{-19}}{6.625 	imes 10^{-34}} \ Hz$.Calculating the result: $
u \approx 1.304 	imes 10^{15} \ Hz$.

$A$ difference of $5.4 \ eV$ 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? (Take $1 \ eV = 1.6 \times 10^{-19} \ J$,$h = 6.625 \times 10^{-34} \ Js$)

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