Question

In a hydrogen atom an electron jumps from the third orbit to the first orbit . Find out the frequency and wavelength of the spectral line ?​

Answer 1

Explanation:

λ

1

=R

H

[

1

2

1

−

3

2

1

]

C=vλ

v=

9

R

H

×C×8

=

9

10967800×3×10

8

×8

Frequency=v=2.925×10

15

Hz.

Answer 2

✏$\sf\red{The\: energy\: of \:the\: photon \:emitted\: shall\: equal}$$\sf\red{ to\: the \:difference\: between\: the\: electron's}$$\sf\red{potential \:energy \:at \:the\: two \:orbits.}$☃️

✏$\sf\underline\purple{The \:Rydberg\: Formula\: suggest \:that\: an\: electron}$$\sf\green{ that \:travels\: (or \:literally \:"falls")\: from\: energy\: level}$$\sf\green{ni\: to \:nf\: (ni \:> \:nf) \:emits\: a \:photon\: of\: wavelength}$$\sf\green{ lamda\: for\: which }$☃️

$\sf{\implies \dfrac{1}{lamda} = R \times (\dfrac{1}{{nf}^{2}} - \dfrac{1}{{ni}^{2}})}$

$\sf{Where \:the\: Ryderberg's \:Constant \:R = 1.097 \times {10}^{7}{m}^{-1}}$

✒$\sf{\underline{\underline{In\: particular\: scenario}}}$

$\sf{\longrightarrow ni = 3}$

$\sf{\longrightarrow nf = 1 }$

✒$\sf{\underline{\underline{Such \:that }}}$

$\sf{\implies \dfrac{1}{lamda} = R \times (\dfrac{1}{{nf}^{2}} - \dfrac{1}{{ni}^{2}})}$

$\sf{= 1.097 \times {10}^{7} {m}^{-1} \times (\dfrac{1}{{1}^{2}} - \dfrac{1}{{3}^{2}})}$

$\sf{= 9.751 \times {10}^{-7} {m}^{-1}}$

$\sf{\implies lamda = 1.026 \times {10}^{-7} \cancel{m} \times \dfrac{{10}^{9}\:nm}{1\:\cancel{m}}}$

$\bold{\fbox{= 102.6 \:nm}}$✍