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

Related Formula

The energy of an electron in the n$n$-th state of a hydrogen atom is:

Core Logic

At T = 0 K$T = 0\text{ K}$, all hydrogen atoms are in their ground state (n = 1$n = 1$). To observe any lines in the Balmer series emission spectrum:

• The emission lines must result from a transition terminating in the state n = 2$n = 2$.
• This requires the hydrogen electron to first be excited to at least the state n = 3$n = 3$.

Thus, the bombarding electrons must have enough kinetic energy to excite the ground-state electron (n = 1$n = 1$) to the n = 3$n = 3$ state.

Step 1: Calculate the Required Energy and Potential

The required energy difference is:

Since the excitation energy is provided by electrons accelerated through potential V$V$, the minimum potential difference required is:

Step 2: Solve for Alpha

Comparing to the given form (α)/(10) V$\frac{\alpha}{10}\text{ V}$:

Pattern Recognition

Balmer series emission requires excitation of the electron to at least n=3$n=3$. Ground state n=1$n=1$ to n=3$n=3$ excitation energy is 12.09 eV ≈ 12.1 eV$12.09\text{ eV} \approx 12.1\text{ eV}$, matching α = 121$\alpha = 121$ when multiplied by 10$10$.

Chapter Mix

Class 12 Physics: Atoms