A hydrogen atom in its ground state absorbs 10.2 eV of energy. The angular momentum of electron of the hydrogen atom will increase by the value of :
(Given, Planck's constant = 6.6 $\times$ 10$-$34 Js).
Solution
<p>$- 13.6 + 10.2 = {{ - 13.6} \over {{n^2}}}$</p>
<p>$\Rightarrow {{13.6} \over {{n^2}}} = 3.4$</p>
<p>$\Rightarrow n = 2$</p>
<p>$$ \Rightarrow \Delta L = 2 \times {h \over {2\lambda }} - 1 \times {h \over {2\lambda }}$$</p>
<p>$= {h \over {2\lambda }}$</p>
<p>$\Rightarrow \Delta L \simeq 1.05 \times {10^{ - 34}}$ Js</p>
About this question
Subject: Physics · Chapter: Atoms and Nuclei · Topic: Bohr's Model of Hydrogen Atom
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