Question

A $12.5eV$ electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

Answer 1

Step 1: Given Data

The energy of the electron beam $E=12.5eV$

Step 2: Calculate the Wavelength

We know that according to quantum mechanics,

$E=\frac{hc}{\lambda }$,

where $h$ is the Planck's constant which is equal to $6.626\times {10}^{-34}kg{m}^2/s$

$c$ is the speed of light and is equal to $3\times {10}^{8}m/s$.

$\therefore \lambda =\frac{hc}{E}$

We know that $1eV=1.6\times {10}^{-19}J$

$\therefore \lambda =\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{12.5\times 1.6\times {10}^{-19}}$

$=993.9\stackrel{\circ }{A}$

Step 3: Estimate the Series

The wavelength for the transition is from $n_1$ to $n_2$ is given by

$\frac{1}{\lambda }=R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$

where, $R$ is the Rydberg's constant which is equal to $1.09\times {10}^7{m}^{-1}$

For the Lyman series, $n_1=1$ and $n_2=\infty$, the shortest wavelength is given as,

$\frac{1}{{\lambda }_s}=R\left(1-0\right)$

$\Rightarrow {\lambda }_s=\frac{1}{R}=912\stackrel{\circ }{A}$

Similarly for the longest wavelength, $n_1=1$ and $n_2=2$, the wavelength is given as

$\frac{1}{{\lambda }_l}=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3R}{4}$

${\lambda }_l=\frac{4}{3R}=1216\stackrel{\circ }{A}$

Therefore the value of $\lambda =993.9\stackrel{\circ }{A}$ lies between ${\lambda }_s$ and ${\lambda }_l$ of the Lyman series.

Hence, the Lyman series of wavelengths will be emitted.