Physics Ncert Solutions Class 12th Physics Class 12th Atoms

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

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Vishal Baghel | Contributor-Level 10

7 months ago

12.9 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 = $\frac{- 13.6}{n^{2}}$ eV

For n = 3, E = $\frac{- 13.6}{9}$ eV = - 1.5 eV

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

During the de-excitation, the electron can jump from n=3 to n=1 d

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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 = $\frac{- 13.6}{n^{2}}$ eV

For n = 3, E = $\frac{- 13.6}{9}$ eV = - 1.5 eV

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

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

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

$\frac{1}{?}$ = $R_{y} (\frac{1}{1^{2}} - \frac{1}{n^{2}})$ where

$R_{y}$ = Rydberg constant = 1.097 $* 10^{7} m^{- 1}$

$? =$ Wavelength of radiation emitted by the transition of electron

For n = 3, we get

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

$? = 1.215 * 10^{- 7} m$ = 102.6 nm

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

$\frac{1}{?}$ = 1.097 $* 10^{7} * (\frac{1}{1^{2}} - \frac{1}{2^{2}})$

$? = 1.215 * 10^{- 7} m$ = 121.5 nm

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

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

$? = 6.563 * 10^{- 7} m$ = 656.3 nm

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

Hence, in Lyman series, two wavelengths i.e. 102.6 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e. 656.3 nm is emitted.

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12.9 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 eVOrbital energy is related to orbit level(n) as:E = <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>13.6</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> eVFor n = 3, E = <math> <mfrac> <mrow> <mrow> <mo>-</mo> <mn>13.6</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </mfrac> </math> eV = - 1.5 eVThis energy is approximately equal to the energy of the gaseous hydrogen. It can be concluded that the electron has jumped from n = 1 to n = 3 level.During the de-excitation, the electron can jump from n=3 to n=1 directly, which forms a line of the Lyman series of hydrogen spectrum.We have the relation for wave number for Lyman series as: <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> </mfrac> </math> = <math> <msub> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mi>y</mi> </mrow> </mrow> </msub> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>-</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mi>n</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> where <math> <msub> <mrow> <mrow> <mi>R</mi> </mrow> </mrow> <mrow> <mrow> <mi>y</mi> <mi> </mi> </mrow> </mrow> </msub> </math> = Rydberg constant = 1.097 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>7</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> <math> <mi>?</mi> <mo>=</mo> <mi> </mi> </math> Wavelength of radiation emitted by the transition of electronFor n = 3, we get <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> </mfrac> </math> = 1.097 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>7</mn> </mrow> </mrow> </msup> <mo>×</mo> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>-</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> <math> <mi>?</mi> <mo>=</mo> <mn>1.215</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> <mi> </mi> <mi>m</mi> </math> = 102.6 nmIf the electron jumps from n = 2 to n = 1, then the wavelength of the radiation is given as: <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> </mfrac> </math> = 1.097 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>7</mn> </mrow> </mrow> </msup> <mo>×</mo> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>-</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> <math> <mi>?</mi> <mo>=</mo> <mn>1.215</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> <mi> </mi> <mi>m</mi> </math> = 121.5 nmIf the transition takes place from n = 3 to n = 2, then the wavelength is given as: <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>?</mi> </mrow> </mrow> </mfrac> </math> = 1.097 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>7</mn> </mrow> </mrow> </msup> <mo>×</mo> <mfenced separators="|"> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>-</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math> <math> <mi>?</mi> <mo>=</mo> <mn>6.563</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> <mi> </mi> <mi>m</mi> </math> = 656.3 nmThis radiation corresponds to the Balmer series of the hydrogen spectrum.Hence, in Lyman series, two wavelengths i.e. 102.6 nm and 121.5 nm are emitted. And in the Balmer series, one wavelength i.e. 656.3 nm is emitted.

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

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