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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

11.36 Compute the typical de Broglie wavelength of an electron in a metal at 27 °C and compare it with the mean separation between two electrons in a metal which is given to be about 2 × 10^–10 m.

[Note: Exercises 11.35 and 11.36 reveal that while the wave-packets associated with gaseous molecules under ordinary conditions are non-overlapping, the electron wave-packets in a metal strongly overlap with one another. This suggests that whereas molecules in an ordinary gas can be distinguished apart, electrons in a metal cannot be distinguished apart from one another. This indistinguishibility has many fundamental implications which you will explore in more advanced Physics courses.]

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Payal Gupta | Contributor-Level 10

5 months ago

11.36 Temperature, T = 27 $?$ = 300 K

Mean separation between two electrons, r = 2 $\times 10^{- 10}$ m

De Broglie wavelength of an electron is given as:

$λ =$ $\frac{h}{\sqrt{3 m k T}}$ , where

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

m = mass of an electron = 9.11 $\times 10^{- 31}$ kg

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

$λ =$ $\frac{6.626 \times 10^{- 34}}{\sqrt{3 \times 9.11 \times 10^{- 31} \times 1.38 \times 10^{- 23} \times 300}}$ = 6.23 $\times 10^{- 9}$ m

Hence, the De Broglie wavelength is much greater than the given inter-electron separation.

11.36 Temperature, T = 27 <math> <mi>?</mi> </math>  = 300 KMean separation between two electrons, r = 2 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> </math>  mDe Broglie wavelength of an electron 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>3</mn> <mi>m</mi> <mi>k</mi> <mi>T</mi> </mrow> </msqrt> </mrow> </mrow> </mfrac> </math> , wherePlanck’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>  Jsm = mass of an electron = 9.11 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>31</mn> </mrow> </mrow> </msup> </math>  kgk = Boltzmann constant = 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> <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>3</mn> <mo>×</mo> <mn>9.11</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>31</mn> </mrow> </mrow> </msup> <mo>×</mo> <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> </msqrt> </mrow> </mrow> </mfrac> </math> = 6.23 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>9</mn> </mrow> </mrow> </msup> </math>  mHence, the De Broglie wavelength is much greater than the given inter-electron separation.

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Given below are two statements:

Statement I : Davission-Germer experiment establishes, the wave nature of electrons.

Statement II : If electrons have wave nature, they can interfere and show diffraction.

In the light of the above statements choose the correct answer from the option given below:

Raj Pandey

Based on theory

Radiation from hydrogen gas excited to first excited state is used for illuminating certain metallic plate. When the same plate is exposed to the radiation from some unknown hydrogen like gas excited to the same level it is found that de-Broglie wavelength of fastest photoelectron has decreased by 2.3 times. It is given that energy corresponding to longest wavelength of Lyman series of the unknown gas is 3 times the ionization energy of hydrogen gas (13.6eV). The work function (in eV) of the metallic plate is ______. [Take (2.3)² = 5.25]

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z² × (13.6) (1 - ¼) = 3 × (13.6)
z = 2 . (i)
h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - Φ
k? = E? - Φ = z²E? - Φ
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A metal exposed to light of wavelength 800nm and emits photoelectrons with a certain kinetic energy. The maximum kinetic energy of photo-electron doubles when light of wavelength 500nm is used. The work function of the metal is: (Take hc = 1230 eV –nm).

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$_{} \frac{}{}$ $K . E_{1} = \frac{1230}{800} - ?$ - (i)

$_{}_{} \frac{}{}$ $K . E_{2} = 2 K . E_{1} = \frac{1230}{500} - ?$ - (ii)

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When light of frequency twice the threshold frequency is incident on the metal plate, the maximum velocity of emitted electron is $υ_{1} .$ When the frequency of incident radiation is increased to five times the threshold value, the maximum velocity of emitted electron becomes $υ_{2} .$ If $υ_{2} = x υ_{1},$ the value of x will be __________.

Vishal Baghel

hu = hu₀ + K.E

Cases u = 2u₀

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K.E₁ = hu₀

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Kindly consider the following

$f (x) = {\begin{matrix} \frac{2 x^{2} - 3 x + 2}{x - 2}, & x \neq 2 \\ 5, & x = 2 \end{matrix}$

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This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} f (x) = \frac{2 x^{2} - 3 x - 2}{x - 2} \\ = \frac{2 x^{2} - 4 x + x - 2}{x - 2} = \frac{2 x (x - 2) + 1 (x - 2)}{x - 2} \\ = \frac{(2 x + 1) (x - 2)}{x - 2} = 2 x + 1 \\ \underset{x \to 2^{-}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 - h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2^{+}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 + h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2}{l i m} f (x) = 5 \\ A s \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) = 5 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 2 . \end{array}$

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