Question

A hydrogen atom is excited from the ground state by an electron beam of energy 12.5 eV. Find the no of lines emmited by atom in excited state

Answer 1

It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV.

Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

Orbital energy is related to orbit level (n) as:

E = -13.6/(n)2 eV

For n = 3, E = -13.6/(9)2 = -1.5 eV

This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

1/λ = Ry (1/12 - 1/n2)

Where,

Ry = Rydberg constant = 1.097 × 107 m−1

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λ as:

1/λ = 1.097 x 107 (1/12 - 1/32)

= 1.097x107(1-1/9) = 1.097x107x8/9

λ = 9/(8 x 1.097 x 107) = 102.55 nm

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

1/λ = 1.097 x 107 (1/12 - 1/22)

= 1.097x107(1-1/4) = 1.097x107x3/4

λ = 4/(1.097x107x3) = 121.54nm

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

1/λ = 1.097 x 107 (1/22 - 1/32)

= 1.097x107(1/4-1/9) = 1.097x107x5/36

λ = 36/(5x1.097x107) = 656.33 nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.

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It is given that the energy of the electron beam used to bombard gaseous hydrogen at room temperature is 12.5 eV.

Also, the energy of the gaseous hydrogen in its ground state at room temperature is −13.6 eV.

When gaseous hydrogen is bombarded with an electron beam, the energy of the gaseous hydrogen becomes −13.6 + 12.5 eV i.e., −1.1 eV.

 

Orbital energy is related to orbit level (n) as:

E = -13.6/(n)2 eV

For n = 3, E = -13.6/(9)2 = -1.5 eV

This energy is approximately equal to the energy of gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.

During its de-excitation, the electrons can jump from n = 3 to n = 1 directly, which forms a line of the Lyman series of the hydrogen spectrum.

We have the relation for wave number for Lyman series as:

1/λ = Ry (1/12 - 1/n2)

Where,

Ry = Rydberg constant = 1.097 × 107 m−1

λ= Wavelength of radiation emitted by the transition of the electron

For n = 3, we can obtain λ as:

1/λ = 1.097 x 107 (1/12 - 1/32)

= 1.097x107(1-1/9) = 1.097x107x8/9

λ = 9/(8 x 1.097 x 107) = 102.55 nm

If the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as:

1/λ = 1.097 x 107 (1/12 - 1/22)

= 1.097x107(1-1/4) = 1.097x107x3/4

λ = 4/(1.097x107x3) = 121.54nm

If the transition takes place from n = 3 to n = 2, then the wavelength of the radiation is given as:

1/λ = 1.097 x 107 (1/22 - 1/32)

= 1.097x107(1/4-1/9) = 1.097x107x5/36

λ = 36/(5x1.097x107) = 656.33 nm

This radiation corresponds to the Balmer series of the hydrogen spectrum.

Hence, in Lyman series, two wavelengths i.e., 102.5 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e., 656.33 nm is emitted.