Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Atoms
0 votes

The energy of an electron in an excited hydrogen atom is $\;-3.4 e V \;.$ Calculate the angular momentum of electron according to Bor's theorey. Plan ck's constant h= $\;6.626\times10^{-34} J-s\;.$

$(a)\;2.11\times10^{-38} J-s \qquad(b)\;2.11\times10^{-31} J-s\qquad(c)\;2.11\times10^{-30} J-s\qquad(d)\;2.11\times10^{-34} J-s$

Can you answer this question?

1 Answer

0 votes
Answer : (d) $\;2.11\times10^{-34} J-s$
Explanation :
The energy of an electron in $\;n^{th}\;$state of hydrozen atom is
Given that $\;\varepsilon _{n}^{H}=-3.4 eV \;,$ hence $\;n^2=\large\frac{-13.6}{-3.4}\;$ or n=2
According to $\;2^{nd}\;$ postulate of Bohr's theory the angular momentum of electron is
$L=\large\frac{n h}{2 \pi}=2 \times\large\frac{6.626\times10^{-34}}{2 \times 3.14}=2.11 \times 10^{-34} J-s$
answered Feb 21, 2014 by yamini.v

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App