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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

11.20 (a) Estimate the speed with which electrons emitted from a heated emitter of an evacuated tube impinge on the collector maintained at a potential difference of 500 V with respect to the emitter. Ignore the small initial speeds of the electrons. The specific charge of the electron, i.e., its e/m is given to be 1.76 × 10¹¹ C kg^–1.

(b) Use the same formula you employ in (a) to obtain electron speed for a collector potential of 10 MV. Do you see what is wrong ? In what way is the formula to be modified?

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Payal Gupta | Contributor-Level 10

4 months ago

11.20 Potential difference across evacuated tube, V = 500 V

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

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ = $e V$

$v = \sqrt{\frac{2 e V}{m}}$ = $\sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 \times 500 \times 1.76 \times 10^{11}}$ = 13.27 $\times 10^{6}$ m/s

Therefore, the speed of each emitted electron is 13.27 $\times 10^{6}$ m/s

Collector potential, V = 10 MV = 10 $\times 10^{6}$ V

The speed is given by $v = \sqrt{\frac{2 e V}{m}}$ $= \sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 \times 10 \times 10^{6} \times 1.76 \times 10^{11}}$ = 1.876 $\times 10^{9}$ 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

...more

11.20 Potential difference across evacuated tube, V = 500 V

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

The speed of each emitted electron is given by the relation of kinetic energy as

$E_{k i n e t i c}$ = $\frac{1}{2} m v^{2}$ = $e V$

$v = \sqrt{\frac{2 e V}{m}}$ = $\sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 \times 500 \times 1.76 \times 10^{11}}$ = 13.27 $\times 10^{6}$ m/s

Therefore, the speed of each emitted electron is 13.27 $\times 10^{6}$ m/s

Collector potential, V = 10 MV = 10 $\times 10^{6}$ V

The speed is given by $v = \sqrt{\frac{2 e V}{m}}$ $= \sqrt{2 V \frac{e}{m}}$ = $\sqrt{2 \times 10 \times 10^{6} \times 1.76 \times 10^{11}}$ = 1.876 $\times 10^{9}$ 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 total energy is given as :

$E = m c^{2}$ , where m = relativistic mass

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

Hence, kinetic energy is given as

$E_{k i n e t i c}$ = m $c^{2}$ - $m_{0} c^{2}$

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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]

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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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