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Dual Nature of Radiation and Matter Ncert Solutions Physics Class 12th Physics Ncert Solutions Class 12th Physics Class 12th Dual Nature of Radiation and Matter

11.35 Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

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Payal Gupta | Contributor-Level 10

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11.35 Room temperature, T = 27 $?$ = 300 K

Atmospheric pressure, P = 1 atm = 1.01 $\times 10^{5}$ Pa

Atomic weight of helium atom = 4

Avogadro’s number, $N_{A}$ = 6.023 $\times 10^{23}$

Boltzmann’s constant, k = 1.38 $\times 10^{- 23}$ J ${m o l}^{- 1} K^{- 1}$

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Average energy of a gas at temperature T is given as:

E = $\frac{3}{2}$ kT = $\frac{3}{2} \times$ 1.38 $\times 10^{- 23} \times 300$ = 6.21 $\times 10^{- 21}$ J

De Broglie wavelength is given as

$λ =$ $\frac{h}{\sqrt{2 m E}}$ , where m = mass of He atom = $\frac{A t o m i c w e i g h t}{A v o g a d r o^{'} s n u m b e r}$ = $\frac{4}{6.023 \times 10^{23}}$

m = 6.641 $\times 10^{- 24}$ gm = 6.641 $\times 10^{- 27}$ kg

$λ =$ $\frac{6.626 \times 10^{- 34}}{\sqrt{2 \times 6.641 \times 10^{- 27} \times 6.21 \times 10^{- 21}}}$ = 7.29 $\times 10^{- 11}$ m

We have ideal gas formula:

PV = RT

PV = kNT

$\frac{V}{N} = \frac{k T}{P}$

Where, V = volume of the gas

N = number of moles of the gas

Mea

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11.35 Room temperature, T = 27 $?$ = 300 K

Atmospheric pressure, P = 1 atm = 1.01 $\times 10^{5}$ Pa

Atomic weight of helium atom = 4

Avogadro’s number, $N_{A}$ = 6.023 $\times 10^{23}$

Boltzmann’s constant, k = 1.38 $\times 10^{- 23}$ J ${m o l}^{- 1} K^{- 1}$

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

Average energy of a gas at temperature T is given as:

E = $\frac{3}{2}$ kT = $\frac{3}{2} \times$ 1.38 $\times 10^{- 23} \times 300$ = 6.21 $\times 10^{- 21}$ J

De Broglie wavelength is given as

$λ =$ $\frac{h}{\sqrt{2 m E}}$ , where m = mass of He atom = $\frac{A t o m i c w e i g h t}{A v o g a d r o^{'} s n u m b e r}$ = $\frac{4}{6.023 \times 10^{23}}$

m = 6.641 $\times 10^{- 24}$ gm = 6.641 $\times 10^{- 27}$ kg

$λ =$ $\frac{6.626 \times 10^{- 34}}{\sqrt{2 \times 6.641 \times 10^{- 27} \times 6.21 \times 10^{- 21}}}$ = 7.29 $\times 10^{- 11}$ m

We have ideal gas formula:

PV = RT

PV = kNT

$\frac{V}{N} = \frac{k T}{P}$

Where, V = volume of the gas

N = number of moles of the gas

Mean separation between two atoms of the gas is given by the relation:

r = $({\frac{V}{N})}^{1 / 3}$ = ${(\frac{k T}{P})}^{1 / 3}$

=( ${\frac{1.38 \times 10^{- 23} \times 300}{1.01 \times 10^{5}})}^{1 / 3}$

= 3.45 $\times 10^{- 9}$ m

Hence, the mean separation between the atoms is much greater than the De Broglie wavelength.

less

11.35 Room temperature, T = 27 <math> <mi>?</mi> </math>  = 300 KAtmospheric pressure, P = 1 atm = 1.01 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </math>  PaAtomic weight of helium atom = 4Avogadro’s number, <math> <msub> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> </mrow> </mrow> </msub> </math>  = 6.023 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </math> Boltzmann’s constant, k = 1.38 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> </math>  J <math> <msup> <mrow> <mrow> <mi>m</mi> <mi>o</mi> <mi>l</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>K</mi> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>1</mn> </mrow> </mrow> </msup> </math> Planck’s constant, h = 6.626 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </math>  JsAverage energy of a gas at temperature T is given as:E = <math> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> </math> kT = <math> <mfrac> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mo>×</mo> </math> 1.38 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>300</mn> </math>  = 6.21 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>21</mn> </mrow> </mrow> </msup> </math> JDe Broglie wavelength is given as <math> <mi>λ</mi> <mo>=</mo> <mi> </mi> </math> <math> <mfrac> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <msqrt> <mrow> <mn>2</mn> <mi>m</mi> <mi>E</mi> </mrow> </msqrt> </mrow> </mrow> </mfrac> </math>  , where m = mass of He atom = <math> <mfrac> <mrow> <mrow> <mi>A</mi> <mi>t</mi> <mi>o</mi> <mi>m</mi> <mi>i</mi> <mi>c</mi> <mi> </mi> <mi>w</mi> <mi>e</mi> <mi>i</mi> <mi>g</mi> <mi>h</mi> <mi>t</mi> </mrow> </mrow> <mrow> <mrow> <mi>A</mi> <mi>v</mi> <mi>o</mi> <mi>g</mi> <mi>a</mi> <mi>d</mi> <mi>r</mi> <msup> <mrow> <mrow> <mi>o</mi> </mrow> </mrow> <mrow> <mrow> <mi>'</mi> </mrow> </mrow> </msup> <mi>s</mi> <mi>n</mi> <mi>u</mi> <mi>m</mi> <mi>b</mi> <mi>e</mi> <mi>r</mi> </mrow> </mrow> </mfrac> </math>  = <math> <mfrac> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> <mrow> <mrow> <mn>6.023</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>23</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> m = 6.641 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>24</mn> </mrow> </mrow> </msup> </math>  gm = 6.641 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>27</mn> </mrow> </mrow> </msup> </math> kg <math> <mi>λ</mi> <mo>=</mo> <mi> </mi> </math> <math> <mfrac> <mrow> <mrow> <mn>6.626</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <msqrt> <mrow> <mn>2</mn> <mo>×</mo> <mn>6.641</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>27</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>6.21</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>21</mn> </mrow> </mrow> </msup> </mrow> </msqrt> </mrow> </mrow> </mfrac> </math> = 7.29 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>11</mn> </mrow> </mrow> </msup> </math> mWe have ideal gas formula:PV = RTPV = kNT <math> <mfrac> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> </mfrac> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>k</mi> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> </mfrac> </math>          Where, V = volume of the gasN = number of moles of the gasMean separation between two atoms of the gas is given by the relation:r = <math> <mo>(</mo> <msup> <mrow> <mrow> <mfrac> <mrow> <mrow> <mi>V</mi> </mrow> </mrow> <mrow> <mrow> <mi>N</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mrow> </msup> </math>   = <math> <msup> <mrow> <mrow> <mo>(</mo> <mfrac> <mrow> <mrow> <mi>k</mi> <mi>T</mi> </mrow> </mrow> <mrow> <mrow> <mi>P</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> =( <math> <msup> <mrow> <mrow> <mfrac> <mrow> <mrow> <mn>1.38</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>23</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>300</mn> </mrow> </mrow> <mrow> <mrow> <mn>1.01</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>/</mo> <mn>3</mn> </mrow> </mrow> </msup> </math> = 3.45 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>9</mn> </mrow> </mrow> </msup> </math> mHence, the mean separation between the atoms is much greater than the De Broglie wavelength.

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11.35 Find the typical de Broglie wavelength associated with a He atom in helium gas at room temperature (27 °C) and 1 atm pressure; and compare it with the mean separation between two atoms under these conditions.

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