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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

11.33 An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de Broglie wavelength associated with the electrons. If other factors (such as numerical aperture, etc.) are taken to be roughly the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light?

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Payal Gupta | Contributor-Level 10

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11.33 The accelerating voltage of the electrons, V = 50 kV = 50 $\times 10^{3}$ V

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

Wavelength of yellow light, $λ$ = 5.9 $\times 10^{- 7}$ m

The kinetic energy of the electron is given as, $E_{k}$ = e $\times V$ = 1.6 $\times 10^{- 19} \times$ 50 $\times 10^{3}$ J = 8 $\times 10^{- 15}$ J

De Broglie wavelength is given by the relation,

$λ = \frac{h}{\sqrt{2 \times m_{e} \times E_{k}}}$ = $\frac{6.626 \times 10^{- 34}}{\sqrt{2 \times 9.11 \times 10^{- 31} \times 8 \times 10^{- 15}}}$ = 5.5 $\times 10^{- 12}$ m

This wavelength is nearly $10^{5}$ times less than the wavelength of a yellow light.

The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nea

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11.33 The accelerating voltage of the electrons, V = 50 kV = 50 $\times 10^{3}$ V

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

Wavelength of yellow light, $λ$ = 5.9 $\times 10^{- 7}$ m

The kinetic energy of the electron is given as, $E_{k}$ = e $\times V$ = 1.6 $\times 10^{- 19} \times$ 50 $\times 10^{3}$ J = 8 $\times 10^{- 15}$ J

De Broglie wavelength is given by the relation,

$λ = \frac{h}{\sqrt{2 \times m_{e} \times E_{k}}}$ = $\frac{6.626 \times 10^{- 34}}{\sqrt{2 \times 9.11 \times 10^{- 31} \times 8 \times 10^{- 15}}}$ = 5.5 $\times 10^{- 12}$ m

This wavelength is nearly $10^{5}$ times less than the wavelength of a yellow light.

The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly $10^{5}$ times that of an optical microscope.

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11.33 The accelerating voltage of the electrons, V = 50 kV = 50 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math>  VMass of the electron, <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>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>  CWavelength of yellow light, <math> <mi>λ</mi> </math> = 5.9 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>7</mn> </mrow> </mrow> </msup> </math> mThe 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>  = e <math> <mo>×</mo> <mi>V</mi> </math>  = 1.6 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mo>×</mo> </math> 50 <math> <mo>×</mo> <mi> </mi> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>3</mn> </mrow> </mrow> </msup> </math>  J = 8 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>15</mn> </mrow> </mrow> </msup> </math> JDe Broglie wavelength is given by the relation, <math> <mi>λ</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>h</mi> </mrow> </mrow> <mrow> <mrow> <msqrt> <mrow> <mn>2</mn> <mo>×</mo> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> </msub> <mo>×</mo> <msub> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mi>k</mi> </mrow> </mrow> </msub> <mi mathvariant="normal"> </mi> </mrow> </msqrt> </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> </mrow> </mrow> <mrow> <mrow> <msqrt> <mrow> <mn>2</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>8</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>15</mn> </mrow> </mrow> </msup> </mrow> </msqrt> </mrow> </mrow> </mfrac> </math>  = 5.5 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>12</mn> </mrow> </mrow> </msup> </math>  mThis wavelength is nearly <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> <mi> </mi> </mrow> </mrow> </msup> </math> times less than the wavelength of a yellow light.The resolving power of a microscope is inversely proportional to the wavelength of light used. Thus, the resolving power of an electron microscope is nearly <math> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>5</mn> <mi> </mi> </mrow> </mrow> </msup> </math> times that of an optical microscope.

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Similar Questions for you

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
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV

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).

Vishal Baghel

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

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

from (i) & (ii)

$? = 0.615$ ev

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

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)

$\frac{v_{2}}{v_{1}} = \sqrt[]{\frac{8}{2} = 2}$

v₂ = 2v₁

Kindly consider the following

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

alok kumar singh

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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