Question

what are the frequency and wavelength of a photon emitted during a transition from n=6 to n=3

Answer 1

Answer:

since n1=5andn1=2, this transition gives rise to a spectral line in the visible region of the balmer series. From equation

ΔE=2.18×10−18J[152−122]

4.58×10−19J

it is an emission energy

the frequency of the photon ( taking energy in terms of magnitude ) is given by

v=ΔEh

4.58×10−19J6.626×10−34Js

6.91×1014Hz

λ=cv=3.0×108ms−16.91×1014Hz=434nm

Explanation:

Answer 2

Answer:

The frequency and wavelength of a photon emitted during a transition from n = 6 to n = 3 is 2.87 × 10¹⁷ Hz and 1.045 nm.

$\qquad$_____________

Explanation:

Given that, the photon is having a transition from :

$\sf\dashrightarrow \: \: n_{1} = 6$

$\sf \dashrightarrow \: \: n_{2} = 3$

$\:$

Firstly, we'll find the energy between the two orbits :

$\sf \dashrightarrow \: \: \triangle E = 2.18 \times {10}^{ - 18} \: J \: \bigg \{ \dfrac{1}{ n_{1} ^{2} } - \dfrac{1}{ n_{ 2 } ^{2} } \bigg \}$

$\sf \dashrightarrow \: \: \triangle E = 2.18 \times {10}^{ - 18} \:J \: \bigg \{ \dfrac{1}{ {(6)}^{2} } - \dfrac{1}{ {(3)}^{2} } \bigg \}$

$\sf \dashrightarrow \: \: \triangle E \: = 2.18 \times {10}^{ - 18} \: J \: \bigg \{ \dfrac{1}{36} - \dfrac{1}{9} \bigg \}$

$\sf \dashrightarrow \: \: \triangle E = 2.18 \times {10}^{ - 18} \: J \: \bigg \{ \dfrac{1 - 4}{36 } \bigg \}$

$\sf \dashrightarrow \: \: \triangle E = 2.18 \times {10}^{- 18} \: J \: \bigg \{ \cancel\cfrac{ - 3}{36} \bigg \}$

$\sf \dashrightarrow \: \: \triangle E = 2.18 \times {10}^{ - 18} \: J \: \bigg \{ \dfrac{ - 1}{12} \bigg \}$

$\sf \dashrightarrow \: \: \triangle E = \dfrac{- 2.18\times 10^{-18}}{12}$

$\sf \dashrightarrow \: \triangle E = 0.18 \times10 ^{ - 18} \: J$

$\:$

Now, let's calculate the frequency of the photon :

$\sf \dashrightarrow \: \: v = \dfrac{ \triangle E}{h}$

$\sf \dashrightarrow~~ v = \dfrac{0.18\times 10^{-18}}{6.626\times 10^{-34}}$

$\sf \dashrightarrow~~ v = 0.287\times 10^{-18-(-34)}$

$\sf \dashrightarrow~~ v = 0.287\times 10^{16}$

$\pmb{\sf{\dashrightarrow~~ v=2.87\times 10^{17}~Hz}}$

$\:$

Finally, calculating the wavelength :

$\sf \dashrightarrow~~ \lambda =\dfrac{c}{v}$

$\sf \dashrightarrow \: \: \lambda = \dfrac{3.0 \times {10}^{8}m {s}^{ - 1} }{2.87 \times {10}^{17} hz}$

$\sf \dashrightarrow \: \: \lambda =1.045 \times {10}^{(8 - 17)}$

$\sf \dashrightarrow \: \: \lambda = 1.045 \times {10}^{ - 9} m$

$\pmb{\sf \dashrightarrow~~ \lambda=1.045 nm}$