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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

11.31 Crystal diffraction experiments can be performed using X-rays, or electrons accelerated through appropriate voltage. Which probe has greater energy? (For quantitative comparison, take the wavelength of the probe equal to 1 Å, which is of the order of inter-atomic spacing in the lattice) ( = 9.11 × 10^–31 kg).

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Payal Gupta | Contributor-Level 10

5 months ago

11.31 Wavelength of the light emitted from probe, $λ$ = 1 Å = 1 $\times 10^{- 10}$ m

Mass of the electron, $m_{e}$ = 9.11 $\times 10^{- 31}$ kg

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

Charge of an electron, e = 1.6 $\times 10^{- 19}$ C

Speed of light, c = 3 $\times 108$
m/s

The kinetic energy of the electron is given as :

$E_{k}$ = $\frac{1}{2} m_{e} v^{2}$ ……………(1)

The de Broglie wavelength is given by

$λ = \frac{h}{p}$ ,where p = momentum = $m_{e} v$

$λ = \frac{h}{m_{e} v}$

$v$ = $\frac{h}{m_{e} λ}$

Substituting the value of v in equation (1), we get

$E_{k}$ = $\frac{1}{2} m_{e} {(\frac{h}{m_{e} λ})}^{2}$ = $\frac{m_{e} h^{2}}{2 {m_{e}}^{2} λ^{2}}$ = $\frac{h^{2}}{2 {m_{e} λ}^{2}}$

= $\frac{{(6.626 \times 10^{- 34})}^{2}}{2 {\times 9.11 \times 10^{- 31} \times (1 \times 10^{- 10})}^{2}}$

= $\frac{2.41 \times 10^{- 17}}{1.6 \times 10^{- 19}}$ ev

= 150.6 eV

Energy of a photon, E’ = $\frac{h c}{λ}$ = $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{1 \times 10^{- 10} \times 1.6 \times 10^{- 19}}$ = 12.42 $\times 10^{3}$ eV= 12.42 keV

Hence, a photon has a greater energy than an electron for the same wavelength.

11.31 Wavelength of the light emitted from probe, <math> <mi>λ</mi> </math> = 1 Å = 1 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> </math> mMass of the electron, <math><msub><mrow><mrow><mi mathvariant="bold-italic">m</mi></mrow></mrow><mrow><mrow><mi mathvariant="bold-italic">e</mi></mrow></mrow></msub></math> = 9.11 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>31</mn> </mrow> </mrow> </msup> </math> kgPlanck’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> JsCharge of an electron, e = 1.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math> CSpeed of light, c = 3 <math><mo>×</mo><mrow><mrow><mn>10</mn></mrow></mrow><mo>8</mo></math>  m/sThe kinetic energy of the electron is given as : <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> </msub> </math> = <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> ……………(1)The de Broglie wavelength is given by <math> <mi>λ</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> </mfrac> </math> ,where p = momentum = <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <mi>v</mi> </math> <math> <mi>λ</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <mi>v</mi> </mrow> </mrow> </mfrac> </math> <math><mi>v</mi><mi></mi></math> = <math><mfrac><mrow><mrow><mi>h</mi></mrow></mrow><mrow><mrow><msub><mrow><mrow><mi>m</mi></mrow></mrow><mrow><mrow><mi>e</mi></mrow></mrow></msub><mi>λ</mi></mrow></mrow></mfrac></math>Substituting the value of v in equation (1), we get <math> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> </msub> </math> = <math> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <msup> <mrow> <mrow> <mo>(</mo> <mfrac> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <mi>λ</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <mfrac> <mrow> <mrow> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <msup> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <msup> <mrow> <mrow> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <msup> <mrow> <mrow> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <mi>λ</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <msup> <mrow> <mrow> <mo>(</mo> <mn>6.626</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>34</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> <msup> <mrow> <mrow> <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> <mo>(</mo> <mn>1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mn>2.41</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>17</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <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> </mfrac> </math> ev= 150.6 eVEnergy of a photon, E’ =  <math> <mfrac> <mrow> <mrow> <mi>h</mi> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> </mfrac> </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> <mo>×</mo> <mn>3</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mn>1</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</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> </mfrac> </math> = 12.42 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math> eV= 12.42 keVHence, a photon has a greater energy than an electron for the same wavelength.

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Given below are two statements:

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

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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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