MCQ
What is the de-Broglie wavelength associated with the hydrogen electron in its third orbit
- A$9.96 \times {10^{ - 10}}\,cm$
- ✓$9.96 \times {10^{ - 8}}\,cm$
- C$9.96 \times {10^4}\,cm$
- D$9.96 \times {10^8}\,cm$
Since,
$\frac{ nh }{2 \pi}= mvr$
$\frac{ h }{ mv }=\frac{2 \pi r }{ n }$
$\lambda=\frac{ h }{ mv }=\frac{2 \pi \times 9 R }{3}$
$\lambda=6 \pi R$
We know that
$R =0.529\, \mathring A$
Therefore,
$\lambda=6 \pi \times 0.529$
$\lambda=9.96\, \mathring A$
$\lambda=9.96 \times 10^{-8} \,cm$
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${P_4} + 10C{l_2}\xrightarrow{\Delta }PC{l_5}$
(Round off to the NearestInteger)
[Use:Atomicmasses:$Na:23.0\,u \mathrm{O}: 16.0 \,\mathrm{u} \quad \mathrm{H}: 1.0 \,\mathrm{u}$, Density of $\mathrm{H}_{2} \mathrm{O}: 1.0 \,\mathrm{~g} \,\mathrm{~cm}^{-3}$ ]
${N_2}{H_4}(l)\,+\,2{H_2}{O_2}(l)\,\, \to \,{N_2}(g)\,\, + \,4{H_2}O(l);\,\,{\Delta _r}H_1^o\, = \, - 818\,\,kJ/mol$
${N_2}{H_4}(l)\,\, + {O_2}(g)\,\, \to \,{N_2}(g)\,\, + \,2{H_2}O(l);\,\,{\Delta _r}H_2^o\, = \, - 622\,\,kJ/mol$
${H_2}(g)\,\, + 1/2{O_2}(g)\,\, \to \,{H_2}O(l);\,\,{\Delta _r}H_3^o\, = \, - 285\,\,kJ/mol$
......$kJ/mol$
Which of the following product $(s)$ is/are can be obtained in the above reaction.