# 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= $\;6.626\times10^{-34} J-s\;.$

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

## 1 Answer

Answer : (d) $\;2.11\times10^{-34} J-s$
Explanation :
The energy of an electron in $\;n^{th}\;$state of hydrozen atom is
$\varepsilon_{n}^{H}=-\large\frac{13.6}{n^2}\;eV$
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

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