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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

11.22 An electron gun with its collector at a potential of 100 V fires out electrons in a spherical bulb containing hydrogen gas at low pressure ( mm of Hg). A magnetic field of 2.83 × 10^–4 T curves the path of the electrons in a circular orbit of radius 12.0 cm. (The path can be viewed because the gas ions in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the ‘fine beam tube’ method.) Determine e/m from the data.

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Payal Gupta | Contributor-Level 10

4 months ago

11.22 Potential, V = 100 V

Magnetic field experienced by electron, B = 2.83 $\times 10^{- 4}$ T

Radius of the circular orbit, r = 12.0 cm = 12 $\times 10^{- 2}$ m

Mass of each electron = m

Charge on each electron = e

Velocity of each electron= v

The energy of each electron is equal to its kinetic energy, i.e.

$\frac{1}{2}$ m $v^{2}$ = $e V$

$v^{2} = \frac{2 e V}{m}$ …….(1)

Since centripetal force ( $\frac{m v^{2}}{r})$ = Magnetic force (evB), we can write

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

$v$ = $\frac{e B r}{m}$ ………………(2)

Equating equations (1) and (2) we get

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

$\frac{e}{m}$ = $\frac{2 V}{B^{2} r^{2}}$ = $\frac{2 \times 100}{({2.83 \times 10^{- 4})}^{2} \times ({12 \times 10^{- 2})}^{2}}$ = 1.734 $\times 10^{11}$ C/kg

Therefore, the specific charge ratio (e/m) is 1.734 $\times 10^{11}$ C/k

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11.22 Potential, V = 100 V

Magnetic field experienced by electron, B = 2.83 $\times 10^{- 4}$ T

Radius of the circular orbit, r = 12.0 cm = 12 $\times 10^{- 2}$ m

Mass of each electron = m

Charge on each electron = e

Velocity of each electron= v

The energy of each electron is equal to its kinetic energy, i.e.

$\frac{1}{2}$ m $v^{2}$ = $e V$

$v^{2} = \frac{2 e V}{m}$ …….(1)

Since centripetal force ( $\frac{m v^{2}}{r})$ = Magnetic force (evB), we can write

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

$v$ = $\frac{e B r}{m}$ ………………(2)

Equating equations (1) and (2) we get

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

$\frac{e}{m}$ = $\frac{2 V}{B^{2} r^{2}}$ = $\frac{2 \times 100}{({2.83 \times 10^{- 4})}^{2} \times ({12 \times 10^{- 2})}^{2}}$ = 1.734 $\times 10^{11}$ C/kg

Therefore, the specific charge ratio (e/m) is 1.734 $\times 10^{11}$ C/kg

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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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