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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

11.21 (a) A monoenergetic electron beam with electron speed of 5.20 × 10⁶ m s ^–1 is subject to a magnetic field of 1.30 × 10^–4 T normal to the beam velocity. What is the radius of the circle traced by the beam, given e/m for electron equals 1.76 × 10¹¹C kg^–1.

(b) Is the formula you employ in (a) valid for calculating radius of the path of a 20 MeV electron beam? If not, in what way is it modified?

[Note: Exercises 11.20(b) and 11.21(b) take you to relativistic mechanics which is beyond the scope of this book. They have been inserted here simply to emphasize the point that the formulas you use in part (a) of the exercises are not valid at very high speeds or energies. See answers at the end to know what ‘very high speed or energy’ means.]

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Payal Gupta | Contributor-Level 10

4 months ago

11.21 Speed of the electron, v = 5.20 $\times 10^{6} m / s$

Magnetic field experienced by the electron, B = 1.30 $\times 10^{- 4}$ T

Specific charge of electron, e/m = 1.76 $\times 10^{11}$ C/kg

Charge of an electron e = 1.60 $\times 10^{- 19}$ C

Mass of electron, m = 9.1 $\times 10^{- 31}$ kg

The force exerted on the electron is given as

F = $e |\overset{?}{v} + \overset{?}{B}|$

= $e v B \sin ? θ$ , where $θ$ = angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam, hence $θ = 90 °$

Therefore, $F = e v B$ ………………(1)

The beam traces a circular path of radius r. The magnetic field due to its bending nature provides a centrifugal forc

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11.21 Speed of the electron, v = 5.20 $\times 10^{6} m / s$

Magnetic field experienced by the electron, B = 1.30 $\times 10^{- 4}$ T

Specific charge of electron, e/m = 1.76 $\times 10^{11}$ C/kg

Charge of an electron e = 1.60 $\times 10^{- 19}$ C

Mass of electron, m = 9.1 $\times 10^{- 31}$ kg

The force exerted on the electron is given as

F = $e |\overset{?}{v} + \overset{?}{B}|$

= $e v B \sin ? θ$ , where $θ$ = angle between the magnetic field and the beam velocity

The magnetic field is normal to the direction of beam, hence $θ = 90 °$

Therefore, $F = e v B$ ………………(1)

The beam traces a circular path of radius r. The magnetic field due to its bending nature provides a centrifugal force (F = $\frac{m v^{2}}{r}$ ) for the beam.

Therefore, $e v B$ = $\frac{m v^{2}}{r}$

r = $\frac{m v^{2}}{e v B}$ = $\frac{m v}{e B}$

= $\frac{9.1 \times 10^{- 31} \times 5.20 \times 10^{6}}{1.60 \times 10^{- 19} \times 1.30 \times 10^{- 4}}$ = 0.2275 m = 22.75 cm

Energy of the electron beam, E = 20 MeV = 20 $\times 10^{6}$ eV = 20 $\times 10^{6} \times 1.6 \times 10^{- 19}$ J

E = 3.2 $\times 10^{- 12}$ J

The energy of the electron beam is given as:

E = $\frac{1}{2}$ m $v^{2}$

v = $\sqrt{\frac{2 E}{m}}$ = $\sqrt{\frac{2 \times 3.2 \times 10^{- 12}}{9.1 \times 10^{- 31}}}$ = 2.652 $\times 10^{9}$ m/s

This is not possible nothing can move faster than the light. In the above formula

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ can only be used in the non-relativistic limit, i.e. v << c

For very high speed problems, relativistic equations must be considered for solving them. In the relativistic limit, the mass is given as :

$m = m_{0} \sqrt{(1 - \frac{v^{2}}{c^{2}}})$ where $m_{0}$ = Mass of the particle at rest.

Hence, the radius of the circular path is given as

r = $\frac{m v}{e B}$ = $\frac{v}{e B} \times m_{0} \sqrt{(1 - \frac{v^{2}}{c^{2}}})$

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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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