JEE Advance - Physics (2015 - Paper 2 Offline - No. 10)
An electron in an excited state of Li2+ ion has angular momentum $${{3h} \over {2\pi }}$$. The de Broglie wavelength of the electron in this state is p$$\pi$$a0 (where a0 is the Bohr radius). The value of p is
Answer
2
Explanation
Angular momentum $$mvr = {{nh} \over {2\pi }}$$ where $$r = 3{a_0}$$ where $$n = 3$$, that is, electron in $$L{i^{2 + }}$$ is in second excited state
$$\lambda = {h \over {mv}} = p\pi {a_0}$$
$$ \Rightarrow n = p\pi (mv{a_0}) = p\pi \left( {{{mvr} \over 3}} \right) = {{p\pi } \over 3}\left( {{{3h} \over {2\pi }}} \right) = {{ph} \over 2}$$
Therefore, $$p = 2$$.
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