MCQ
If the de Brogile wavelength of the electron in $n^{th}$ Bohr orbit in a hydrogenic atom is equal to $1.5\,\pi a_0$ ( $a_0$ is Bohr radius), then the value of $n/z$ is
- A$0.40$
- B$1.50$
- C$1.0$
- ✓$0.75$
$\lambda = \frac{{2\pi r}}{n} = \frac{{2\pi {n^2}{a_0}}}{{n \times Z}} = 2\pi \frac{n}{Z}{a_0}$
$\lambda = 1.5\pi {a_0}$
$\therefore \,2\pi \frac{n}{Z}{a_0} = 1.5\pi {a_0}$
$\therefore \frac{n}{Z} = \frac{{1.5}}{2} = 0.75$
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$CS_2 + 3O_2 \to 2SO_2 + CO_2$