11.34 The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10–15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)

11.34 Wavelength of a proton or a neutron, λ ≈ 10 - 15 mRest mass energy of an electron: m 0 c 2 = 0.511 MeV = 0.511 × 10 6 eV= 0.511 × 10 6 × 1.6 × 10 - 19 J m 0 c = 8.176 × 10 - 14 JPlanck’s constant, h = 6.626 × 10 - 34 JsSpeed of light, c = 3 × 10 8 m/sThe momentum of a proton or a neutron is given as: p = h λ = 6.626 × 10 - 34 10 - 15 = 6.626 × 10 - 19 kgm/sThe relativistic relation for energy (e) is given as: E 2 = p 2 c 2 + m 0 2 c 2 = ( 6.626 × 10 - 19 ) 2 × ( 3 × 10 8 ) 2 + ( 8.176 × 10 - 14 ) 2 = 4.390 × 10 - 37 × 9 × 10 16 + 6.684 × 10 - 27 =3.951 × 10 - 20 E = 1.988 × 10 - 10 J = 1.988 × 10 - 10 1.6 × 10 - 19 eV = 1.242 × 10 9 eV = 1.242 BeV

Ask & Answer Question

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

11.34 The wavelength of a probe is roughly a measure of the size of a structure that it can probe in some detail. The quark structure of protons and neutrons appears at the minute length-scale of 10^–15 m or less. This structure was first probed in early 1970’s using high energy electron beams produced by a linear accelerator at Stanford, USA. Guess what might have been the order of energy of these electron beams. (Rest mass energy of electron = 0.511 MeV.)

2 Views | Posted 5 months ago

Asked by Shiksha User

1 Answer

Answered by

Payal Gupta | Contributor-Level 10

5 months ago

11.34 Wavelength of a proton or a neutron, $λ \approx 10^{- 15}$ m

Rest mass energy of an electron: $m_{0} c^{2}$ = 0.511 MeV = 0.511 $\times 10^{6}$ eV

= 0.511 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

$m_{0} c$ = 8.176 $\times 10^{- 14}$ J

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

Speed of light, c = 3 $\times 10^{8}$ m/s

The momentum of a proton or a neutron is given as:

$p$ = $\frac{h}{λ}$ = $\frac{6.626 \times 10^{- 34}}{10^{- 15}}$ = 6.626 $\times 10^{- 19}$ kgm/s

The relativistic relation for energy (e) is given as:

$E^{2}$ = $p^{2} c^{2}$ + $m_{0}^{2} c^{2}$

= ${(6.626 \times 10^{- 19})}^{2} \times {(3 \times 10^{8})}^{2}$ + ${(8.176 \times 10^{- 14})}^{2}$

= 4.390 $\times 10^{- 37} \times 9 \times 10^{16}$ + 6.684 $\times 10^{- 27}$

=3.951 $\times 10^{- 20}$

E = 1.988 $\times 10^{- 10}$ J = $\frac{1.988 \times 10^{- 10}}{1.6 \times 10^{- 19}}$ eV = 1.242 $\times 10^{9}$ eV = 1.242 BeV

11.34 Wavelength of a proton or a neutron, <math> <mi>λ</mi> <mo>≈</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>15</mn> </mrow> </mrow> </msup> </math> mRest mass energy of an electron: <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = 0.511 MeV = 0.511 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</mn> </mrow> </mrow> </msup> </math>  eV= 0.511 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>6</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> </math>  J <math> <msub> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> </msub> <mi>c</mi> <mi> </mi> </math> = 8.176 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>14</mn> </mrow> </mrow> </msup> </math> JPlanck’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> JsSpeed of light, c = 3 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </msup> </math>  m/sThe momentum of a proton or a neutron is given as: <math> <mi>p</mi> </math> = <math> <mfrac> <mrow> <mrow> <mi>h</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> </mrow> </mrow> <mrow> <mrow> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>15</mn> </mrow> </mrow> </msup> </mrow> </mrow> </mfrac> </math>  = 6.626 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> </math>  kgm/sThe relativistic relation for energy (e) is given as: <math> <msup> <mrow> <mrow> <mi>E</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <msup> <mrow> <mrow> <mi>p</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> + <math> <msubsup> <mrow> <mrow> <mi>m</mi> </mrow> </mrow> <mrow> <mrow> <mn>0</mn> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msubsup> <msup> <mrow> <mrow> <mi>c</mi> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = <math> <msup> <mrow> <mrow> <mo>(</mo> <mn>6.626</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>19</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> <mo>×</mo> <msup> <mrow> <mrow> <mo>(</mo> <mn>3</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>8</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> + <math> <msup> <mrow> <mrow> <mo>(</mo> <mn>8.176</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>14</mn> </mrow> </mrow> </msup> <mo>)</mo> </mrow> </mrow> <mrow> <mrow> <mn>2</mn> </mrow> </mrow> </msup> </math> = 4.390 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>37</mn> </mrow> </mrow> </msup> <mo>×</mo> <mn>9</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>16</mn> </mrow> </mrow> </msup> </math>  + 6.684 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>27</mn> </mrow> </mrow> </msup> </math> =3.951 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>20</mn> </mrow> </mrow> </msup> </math> E = 1.988 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</mn> </mrow> </mrow> </msup> </math>  J = <math> <mfrac> <mrow> <mrow> <mn>1.988</mn> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mo>-</mo> <mn>10</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 = 1.242 <math> <mo>×</mo> <msup> <mrow> <mrow> <mn>10</mn> </mrow> </mrow> <mrow> <mrow> <mn>9</mn> </mrow> </mrow> </msup> </math>  eV = 1.242 BeV

0 Upvotes 0 Downvotes

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:

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

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

65k Colleges
1.2k Exams
682k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

1 Answer

Similar Questions for you

Learn more about...

Most viewed information

Most viewed information

Share Your College Life Experience

Didn't find the answer you were looking for?

Need guidance on career and education? Ask our experts

Your Question