Question 321: Atoms & Nuclei

Question

Hydrogen atom is bombarded with electrons accelerated through a potential difference of $$\mathrm{V}$$, which causes excitation of hydrogen atoms. If the experiment is being performed at $$\mathrm{T}=0 \mathrm{~K}$$, the minimum potential difference needed to observe any Balmer series lines in the emission spectra will be $$\frac{\alpha}{10} \mathrm{~V}$$, where $$\alpha=$$ __________.

Accepted Answer

For minimum potential difference electron has to make transition from $$n=3$$ to $$n=2$$ state but first electron has to reach to $$\mathrm{n}=3$$ state from ground state. So, energy of bombarding electron should be equal to energy difference of $$\mathrm{n=3}$$ and $$\mathrm{n=1}$$ state.
$$\begin{aligned}
& \Delta \mathrm{E}=13.6\left[1-\frac{1}{3^2}ight] \mathrm{e}=\mathrm{eV} \
& \frac{13.6 	imes 8}{9}=\mathrm{V} \
& \mathrm{V}=12.09 \mathrm{~V} \approx 12.1 \mathrm{~V}
\end{aligned}$$
So, $$\alpha=121$$

Atoms & Nuclei

Integer / Numeric Answer Type