Question

A 12.5 eV electron beam is used to excite a gaseous hydrogen atom at room temperature. Determine the wavelengths and corresponding series of the lines emitted.

Answer 1

The wavelength of emitted series is 99.2 nm and this belongs to the Lyman series of Bohr's Hydrogen Spectrum.

The wavelength of series emitted during transition is given by the formula:

hc / E = l

E = 12.5 eV
hc = 1240 eV

Therefore wavelength of emitted series is:

l = 1240 / 12.5 = 99.2 nm
This belongs to the Lyman series of Bohr's Hydrogen Spectrum.

Answer 2

Answer:

E is inversely proportional to n^2

Energy levels of hydrogen,

-13.6, -3.4,-1.5

Therefore a 12.5 electron beam can excite a gaseous hydrogen to n=3.

Therefore the corresponding wavelengths are,

102nm, 122nm and 653nm.

Explanation:

It can excite from energy level,

1) 3 ------ 1

2) 2 ------1

3) 3 -------2

Therefore the corresponding wavelengths are,

1) 3 ----- 1 = ( -1.5+13.6) = 12.1 ev

Now wavelength= 1240/12.1 = 102nm

2) 2------1 = ( -3.4+13.6) = 10.2 ev

Now wavelength = 1240/10.2= 122nm

3) 3-----2 = (-1.5+3.4) = 1.9 ev

Now wavelength=1240/ 1.9 = 653nm

102nm,122nm lies in Lyman series.

653nm lies in brackett series.