logo

Ask Questions, Get Answers

X
 
Home  >>  JEEMAIN and NEET  >>  Chemistry  >>  Atomic Structure

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 $h=6.62\times 10^{-34}Jsec$

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

2 Answers

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.
answered Jan 15, 2014 by sreemathi.v
 
∆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.
answered Jul 6, 2016 by mandarkulkarni11
 

Related questions

Download clay6 mobile appDownload clay6 mobile app
...
X