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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

11.37 Answer the following questions:

(a) Quarks inside protons and neutrons are thought to carry fractional charges [(+2/3)e ; (–1/3)e]. Why do they not show up in Millikan’s oil-drop experiment?

(b) What is so special about the combination e/m? Why do we not simply talk of e and m separately?

(c) Why should gases be insulators at ordinary pressures and start conducting at very low pressures?

(d) Every metal has a definite work function. Why do all photoelectrons not come out with the same energy if incident radiation is monochromatic? Why is there an energy distribution of photoelectrons?

(e) The energy and momentum of an electron are related to the frequency and wavelength of the associated matter wave by the relations:

E = h $ν$ , p = $\frac{h}{λ}$ , But while the value of is physically significant, the value of (and therefore, the value of the phase speed ) has no physical significance. Why?

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Payal Gupta | Contributor-Level 10

5 months ago

11.37 (a) Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.

(b) The basic relations for electric field and magnetic field are

$\{e V = \frac{1}{2} m v^{2}\}$ and $\{e B V = \frac{m v^{2}}{r}\}$ respectively

These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius) and B (magnetic field. These relations give the value of the velocity of an electron as

$(v = \sqrt{2 V (\frac{e}{m})}) a n d (v = B r (\frac{e}{m}))$ respectively

It can be observed from these relations that the dynamics of an electron is de

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(b) The basic relations for electric field and magnetic field are

$\{e V = \frac{1}{2} m v^{2}\}$ and $\{e B V = \frac{m v^{2}}{r}\}$ respectively

These relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius) and B (magnetic field. These relations give the value of the velocity of an electron as

$(v = \sqrt{2 V (\frac{e}{m})}) a n d (v = B r (\frac{e}{m}))$ respectively

It can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio of . $\frac{e}{m}$

(c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. A low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.

(d) The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions

(e) The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength( $λ)$ ( is significant, but the frequency( $ν)$ ( associated with an electron has no direct physical significance.

Therefore, the product $ν λ$ (phase speed) has no physical significance.

Group speed is given as:

$v_{G}$ = $\frac{d v}{d k}$ = $\frac{d v}{d (\frac{1}{λ})}$ = dE/dp = (d(p^2/2m))/dp = p/m

This quantity has a physical meaning.

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11.37 (a) Quarks inside protons and neutrons carry fractional charges. This is because nuclear force increases extremely if they are pulled apart. Therefore, fractional charges may exist in nature; observable charges are still the integral multiple of an electrical charge.(b) The basic relations for electric field and magnetic field are <math> <mfenced open="{" close="}" separators="|"> <mrow> <mrow> <mi>e</mi> <mi>V</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </mfrac> <mi>m</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfenced> </math> and <math> <mfenced open="{" close="}" separators="|"> <mrow> <mrow> <mi>e</mi> <mi>B</mi> <mi>V</mi> <mo>=</mo> <mi> </mi> <mfrac> <mrow> <mrow> <mi>m</mi> <msup> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </mrow> </mrow> <mrow> <mrow> <mi>r</mi> </mrow> </mrow> </mfrac> </mrow> </mrow> </mfenced> </math>  respectivelyThese relations include e (electric charge), v (velocity), m (mass), V (potential), r (radius) and B (magnetic field. These relations give the value of the velocity of an electron as <math> <mfenced separators="|"> <mrow> <mrow> <mi>v</mi> <mo>=</mo> <mi> </mi> <msqrt> <mrow> <mn>2</mn> <mi>V</mi> <mo>(</mo> <mfrac> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </msqrt> </mrow> </mrow> </mfenced> <mi> </mi> <mi>a</mi> <mi>n</mi> <mi>d</mi> <mi> </mi> <mfenced separators="|"> <mrow> <mrow> <mi>v</mi> <mo>=</mo> <mi>B</mi> <mi>r</mi> <mo>(</mo> <mfrac> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> </mfenced> </math> respectivelyIt can be observed from these relations that the dynamics of an electron is determined not by e and m separately, but by the ratio of  . <math> <mfrac> <mrow> <mrow> <mi>e</mi> </mrow> </mrow> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> </mfrac> </math> (c) At atmospheric pressure, the ions of gases have no chance of reaching their respective electrons because of collision and recombination with other gas molecules. Hence, gases are insulators at atmospheric pressure. A low pressures, ions have a chance of reaching their respective electrodes and constitute a current. Hence, they conduct electricity at these pressures.(d) The work function of a metal is the minimum energy required for a conduction electron to get out of the metal surface. All the electrons in an atom do not have the same energy level. When a ray having some photon energy is incident on a metal surface, the electrons come out from different levels with different energies. Hence, these emitted electrons show different energy distributions(e) The absolute value of energy of a particle is arbitrary within the additive constant. Hence, wavelength( <math> <mi>λ</mi> <mo>)</mo> </math> ( is significant, but the frequency( <math> <mi>ν</mi> <mo>)</mo> </math> (  associated with an electron has no direct physical significance.Therefore, the product <math> <mi>ν</mi> <mi>λ</mi> </math>  (phase speed) has no physical significance.Group speed is given as: <math> <msub> <mrow> <mrow> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>G</mi> </mrow> </mrow> </msub> </math> = <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mi>k</mi> </mrow> </mrow> </mfrac> </math> = <math> <mfrac> <mrow> <mrow> <mi>d</mi> <mi>v</mi> </mrow> </mrow> <mrow> <mrow> <mi>d</mi> <mo>(</mo> <mfrac> <mrow> <mrow> <mn>1</mn> </mrow> </mrow> <mrow> <mrow> <mi>λ</mi> </mrow> </mrow> </mfrac> <mo>)</mo> </mrow> </mrow> </mfrac> </math> = dE/dp = (d(p^2/2m))/dp = p/mThis quantity has a physical meaning.

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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]

Vishal Baghel

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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