# A 12.9 Ev Beam of Electronic is Used to Bombard Gaseous Hydrogen at Room Temperature. Upto Which Energy Level the Hydrogen Atoms Would Be Excited ? - Physics

A 12.9 eV beam of electronic is used to bombard gaseous hydrogen at room temperature. Upto which energy level the hydrogen atoms would be excited ?
Calculate the wavelength of the first member of Paschen series and first member of Balmer series.

#### Solution

Energy of the electron in the nth state of an atom = $- \frac{13 . 6 z^2}{n^2}$ eV
Here, z is the atomic number of the atom.
For hydrogen atom, z = 1

Energy required to excite an atom from initial state (ni) to final state (nf)  = $- \frac{13 . 6}{{n_f}^2} + \frac{13 . 6}{{n_i}^2}$eV
This energy must be equal to or less than the energy of the incident electron beam.

∴ $- \frac{13 . 6}{{n_f}^2} + \frac{13 . 6}{{n_i}^2}$ 12.9
Energy of the electron in the ground state  = ∴  $- \frac{13 . 6}{1^2} = - 13 . 6$  eV

$\therefore - \frac{13 . 6}{{n_f}^2} + 13 . 6 = 12 . 9$

$13 . 6 - 12 . 9 = \frac{13 . 6}{{n_f}^2}$

${n_f}^2 = \frac{13 . 6}{0 . 7} = 19 . 43$

$\Rightarrow n_f = 4 . 4$

State cannot be a fraction number.
∴ $n_f$ =4

Hence, the hydrogen atom would be excited up to

$4^{th}$ energy level.
Rydberg's formula for the spectrum of the hydrogen atom is given by:
$\frac{1}{\lambda} = R\left[ \frac{1}{{n_1}^2} - \frac{1}{{n_2}^2} \right]$
Here,
$\lambda$  is the wavelength.
Rydberg's constant, R = $1 . 097 \times {10}^7$ m-1
For the first member of the Paschen series:

$n_1 = 3$

$n_2 = 4$

$\frac{1}{\lambda} = 1 . 097 \times {10}^7 \left[ \frac{1}{3^2} - \frac{1}{4^2} \right]$

$\lambda = 18761$Å

For the first member of Balmer series:

$n_1 = 2$

$n_2 = 3$

$\frac{1}{\lambda} = 1 . 097 \times {10}^7 \left[ \frac{1}{2^2} - \frac{1}{3^2} \right]$

$\lambda = 6563$Å

Concept: Energy Levels
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