# Q 12.9  A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?

S Sayak

Since the energy of the electron beam is 12.5 eV the Hydrogen atoms will get excited to all requiring energy equal to or less than 12.5 eV

E=-13.6 eV

E= -1.5 eV

E3 -E1 = 12.1 eV

E4= -0.85 eV

E4-E1=12.75 eV

Therefore the electron can reach maximum upto the level n=3.

During de-excitations, the electron can jump directly from n=3 to n=1 or it can first jump from n=3 to n=2 and then from n=2 to n=1

Therefore two wavelengths from the Lyman series and one from the Balmer series will be emitted

To find the wavelengths emitted we will use the Rydberg's Formula

$\frac{1}{\lambda }=R(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}})$     where R is the Rydberg's constant and equals 1.097$\times$107 m-1

For n1=1 and n2=3

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{1^{2}}-\frac{1}{3^{2}})$

Emitted wavelength is 102.5 nm

For n1=1 and n2=2

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{1^{2}}-\frac{1}{2^{2}})$

Emitted wavelength is 121.54 nm

For n1=2 and n2=3

$\frac{1}{\lambda }=1.097\times 10^{7}(\frac{1}{2^{2}}-\frac{1}{3^{2}})$

Emitted wavelength is 656.3 nm

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