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

11.1 Find the

(a) Maximum frequency, and

(b) Minimum wavelength of X-rays produced by 30 kV electrons.

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Answered by

Payal Gupta | Contributor-Level 10

4 months ago

11.1 Potential of the electrons, V = 30 kV = 3 $\times 10^{4}$ V

Energy of the electron, E = e $\times V$ where e = charge of an electron = 1.6 $\times 10^{- 19} C$

(a) Maximum frequency produced by the X-ray = $ν$

The energy of the electron is given by the relation, E = h $ν$ ,

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

$ν = \frac{E}{h}$ = $\frac{3 \times 10^{4} \times 1.6 \times 10^{- 19}}{6.626 \times 10^{- 34}}$ = 7.244 $\times 10^{18}$ Hz

$H e n c e t h e m a x i u m f r e q u e n c y o f X - r a y p r o d u c e d i s$ 7.244 $\times 10^{18}$ Hz

(b) The minimum wavelength produced is given as

$λ = \frac{c}{ν}$ where c = Speed of light in air, c = 3 $\times 10^{18}$ m/s

$λ = \frac{3 \times 10^{8}}{7.244 \times 10^{18}}$ = 4.14 $\times 10^{- 11}$ m = 0.0414 nm

Hence, the minimum wavelength produced is 0.414 nm.

11.1 Potential of the electrons, V = 30 kV = 3 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> </math>  VEnergy of the electron, E = e <math> <mo>×</mo> <mi>V</mi> </math>  where e = charge of an electron = 1.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mi>C</mi> </math> (a) Maximum frequency produced by the X-ray = <math> <mi>ν</mi> </math> The energy of the electron is given by the relation, E = h <math> <mi>ν</mi> </math> ,where h = Planck’s constant = 6.626 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> </math>  Js <math> <mi>ν</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>3</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>4</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>1.6</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </mrow> </mrow> <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> </mfrac> </math>  = 7.244 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>18</mn> </mrow> </mrow> </msup> </math>  Hz <math> <mi mathvariant="normal">H</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">t</mi> <mi mathvariant="normal">h</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">x</mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">u</mi> <mi mathvariant="normal">m</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">q</mi> <mi mathvariant="normal">u</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">n</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">y</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">f</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">X</mi> <mo>-</mo> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">a</mi> <mi mathvariant="normal">y</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">p</mi> <mi mathvariant="normal">r</mi> <mi mathvariant="normal">o</mi> <mi mathvariant="normal">d</mi> <mi mathvariant="normal">u</mi> <mi mathvariant="normal">c</mi> <mi mathvariant="normal">e</mi> <mi mathvariant="normal">d</mi> <mi mathvariant="normal"> </mi> <mi mathvariant="normal">i</mi> <mi mathvariant="normal">s</mi> <mi mathvariant="normal"> </mi> </math> 7.244 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>18</mn> </mrow> </mrow> </msup> </math>  Hz (b) The minimum wavelength produced is given as  <math> <mi>λ</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mi>ν</mi> </mrow> </mrow> </mfrac> </math> where c = Speed of light in air, c = 3 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>18</mn> </mrow> </mrow> </msup> </math>  m/s  <math> <mi>λ</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mn>3</mn> <mi mathvariant="normal"> </mi> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>7.244</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>18</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = 4.14 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>11</mn> </mrow> </mrow> </msup> </math>  m = 0.0414 nmHence, the minimum wavelength produced is 0.414 nm.

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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:

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

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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 __________.

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hu = hu₀ + K.E

Cases u = 2u₀

h2u₀ = hu₀ + K.E₁

K.E₁ = hu₀

$\frac{1}{2} m v_{1}^{2} = h υ_{0}$

$v_{1}^{2} = \frac{2 h υ_{0}}{m}$

$v_{1} = \sqrt[]{\frac{2 h υ_{0}}{m}}$ - (1)

Now, cases 2

h 5u₀ = hu₀ + k.E₂

k.E₂ = 4hu₀

$\frac{1}{2} m v_{2}^{2} = 4 h υ_{0}$

v₂ = $\sqrt[]{\frac{8 h υ_{0}}{m}}$ - (2)

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v₂ = 2v₁

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

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

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11.1 Find the

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(b) Minimum wavelength of X-rays produced by 30 kV electrons.

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11.1 Find the

(a) Maximum frequency, and

(b) Minimum wavelength of X-rays produced by 30 kV electrons.