MCQ
The energy of an electron in excited hydrogen atom is $-3.4 \,eV.$ Then, according to Bohr's theory the angular momentum of this electron in $Jsec$ is
- ✓$2.1 × 10^{-34}$
- B$4.2 × 10^{ -34}$
- C$6.6 × 10^{-34}$
- D$3.2 × 10^{-33}$
$ \mathrm{E}_{\mathrm{n}}=\frac{-13.6}{\mathrm{n}^{2}} \Rightarrow \mathrm{n}^{2}=\frac{-13.6}{\mathrm{E}_{\mathrm{n}}} $
$ \Rightarrow \mathrm{n}^{2} =\frac{-13.6}{-3.4}=4 $
$ \Rightarrow \mathrm{n}= 2$
$ \Rightarrow \text { angular momentum } =\frac{\mathrm{nh}}{2 \pi}=\frac{2 \times \mathrm{h}}{2 \pi}=\frac{\mathrm{h}}{\pi} $
$=\frac{6.6 \times 10^{-4}}{3.14} $
$=2.1 \times 10^{-34} \mathrm\,{Jsec} $
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[Use : $\left.\mathrm{h}=6.63 \times 10^{-34}\, \mathrm{Js}, \mathrm{m}_{\mathrm{e}}=9.0 \times 10^{-31}\, \mathrm{~kg}\right]$
$\mathrm{X} \rightleftharpoons \mathrm{Y} ; \mathrm{K}_1=1.0$
$\mathrm{Y} \rightleftharpoons \mathrm{Z} ; \mathrm{K}_2=2.0$
$\mathrm{Z} \rightleftharpoons \mathrm{W} ; \mathrm{K}_3=4.0$
The equilibrium constant for the reaction $\mathrm{X} \rightleftharpoons \mathrm{W}$ is