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Calculate the energy emitted when electrons of 1.0g atom of hydrogen under go transition giving the spectral lines of lowest energy in the visible region of its atomic spectra.$R_H=1.1\times 10^7m^{-1},c=3\times 10^8msec^{-1}$ and $h=6.62\times 10^{-34}Jsec$

$(a)\;182.5KJ\qquad(b)\;205KJ\qquad(c)\;150.89KJ\qquad(d)\;190.36KJ$

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A)
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For visible line spectrum,(i.e) Balmer series $n_1=2$.Also for minimum energy transition $n_2=3$
$\large\frac{1}{\lambda}=$$R_H\bigg[\large\frac{1}{n_1^2}-\frac{1}{n_2^2}\bigg]$
$\large\frac{1}{\lambda}=$$R_H\bigg[\large\frac{1}{2^2}-\frac{1}{3^2}\bigg]$
$\Rightarrow 1.1\times 10^7\big[\large\frac{1}{4}-\frac{1}{9}\big]$
$\Rightarrow 1.1\times 10^7\times \large\frac{5}{36}$
$\lambda=6.55\times 10^{-7}m$
Now $E=\large\frac{hc}{\lambda}$
$\Rightarrow \large\frac{6.62\times 10^{-34}\times 3\times 10^8}{6.55\times 10^{-7}}$
$\Rightarrow 3.03\times 10^{-19}J$
If N electrons show this transition in 1g atom of H then
Energy released=$E\times N$
$\Rightarrow 3.03\times 10^{-19}\times 6.023\times 10^{23}$
$\Rightarrow 18.25\times 10^4$
$\Rightarrow 182.5KJ$
Hence (a) is the correct answer.
 
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∆e=13.6 z²[1/n1²-1/n2²] For 1 mole, 6*10²³ * 13.6 * 5/36. Z=1(atomic no. of H = 1) 182.5 kj. n1= 2, n2= 4 Simple method by MANDAR Comment if liked it.
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