Question

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?

Answer 1

Hey Dear,

◆ Answer -

ν = 5.55×10^14 Hz

● Explaination -

# Given -

∆E = 2.3 eV = 2.3×1.6×10^-19 V

# Solution -

Energy of transition of electron is calculated by -

∆E = h.ν

ν = ∆E/h

ν = 2.3×1.6×10^-19 / 6.63×10^-34

ν = 5.55×10^14 Hz

Therefore, radiation of frequency 5.55×10^14 Hz will be emitted.

Thanks dear...

Answer 2

Explanation:

$\Large{\red{\underline{\underline{\sf{\blue{Solution:}}}}}}$

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Separation of two energy levels in an atom,

$\tt E\:=\:2.3\,eV$

$\tt E\:=\:2.3\times 1.6\times 10^{-19}J$

$\tt E\:=\:3.68\times 10^{-19}J$

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$\hookrightarrow$ Let v be the frequency of radiation emitted when the atom transits from the upper level to lower level.

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We have relation for energy as

$\longrightarrow$ $\tt E\:=\:hv$

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Where,

$\hookrightarrow$ $\texttt{h = plank's constant =}$ $\tt 6.62\times 10^{-34}Js$

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$\tt \therefore\:v\:=\:\dfrac{E}{h}$

 ‎ ‎ ‎ ‎ ‎ ‎ ‎

$\tt \implies\:v\:=\: \dfrac{3.68\times 10^{-19}}{6.62\times 10^{-32}}$

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$\tt v\:=\:5.55\times 10^{14}Hz$

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$\hookrightarrow$ Hence the frequency of the radiation is $\tt{\pink{v\:=\:5.55\times 10^{14}Hz}}$