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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

11.2 The work function of cesium metal is 2.14 eV. When light of frequency 6 $\times 10^{14}$ Hz is incident on the metal surface, photoemission of electrons occurs. What is the

(a) Maximum kinetic energy of the emitted electrons,

(b) Stopping potential, and

(c) Maximum speed of the emitted photoelectrons?

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Payal Gupta | Contributor-Level 10

4 months ago

11.2 Work function of cesium metal, $\emptyset_{0}$ = 2.14 eV

Frequency of light, $ν$ = 6 $\times 10^{14}$ Hz

The maximum kinetic energy is given by the photoelectric effect, $K$ = $h ν - \emptyset_{0}$ ,

Where $h$ = Planck’s constant = 6.626 $\times 10^{- 34}$ Js, 1 eV = 1.602 $\times 10^{- 19}$ J

$K$ = $\frac{6.626 \times 10^{- 34} \times 6 \times 10^{14}}{1.602 \times 10^{- 19}} - 2.14$ = 0.345 eV

$H e n c e t h e m a x i m u m k i n e t i c e n e r g y o f t h e e m i t t e d e l e c t r o n i s 0.3416 e V$

For stopping potential $V_{0}$ , we can write the equation for kinetic energy as:

$K$ = $V_{0}$ , where e = charge of an electron = $1.6 \times 10^{- 19}$

or $V_{0} = \frac{K}{e}$ = $\frac{0.345 \times 1.602 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ = 0.345 V

Hence, the stopping potential is 0.345 V

Maximum speed of the emitted photoelectrons = v

Kinetic energy K = $\frac{1}{2}$ m $v^{2}$ , where m = mass of

...more

11.2 Work function of cesium metal, $\emptyset_{0}$ = 2.14 eV

Frequency of light, $ν$ = 6 $\times 10^{14}$ Hz

The maximum kinetic energy is given by the photoelectric effect, $K$ = $h ν - \emptyset_{0}$ ,

Where $h$ = Planck’s constant = 6.626 $\times 10^{- 34}$ Js, 1 eV = 1.602 $\times 10^{- 19}$ J

$K$ = $\frac{6.626 \times 10^{- 34} \times 6 \times 10^{14}}{1.602 \times 10^{- 19}} - 2.14$ = 0.345 eV

$H e n c e t h e m a x i m u m k i n e t i c e n e r g y o f t h e e m i t t e d e l e c t r o n i s 0.3416 e V$

For stopping potential $V_{0}$ , we can write the equation for kinetic energy as:

$K$ = $V_{0}$ , where e = charge of an electron = $1.6 \times 10^{- 19}$

or $V_{0} = \frac{K}{e}$ = $\frac{0.345 \times 1.602 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ = 0.345 V

Hence, the stopping potential is 0.345 V

Maximum speed of the emitted photoelectrons = v

Kinetic energy K = $\frac{1}{2}$ m $v^{2}$ , where m = mass of the electron = 9.1 $\times 10^{- 31}$ kg

Hence, v = $\sqrt{\frac{2 K}{m}}$ = $\sqrt{\frac{2 \times 0.345 \times 1.602 \times 10^{- 19}}{9.1 \times 10^{- 31}}}$ = 348.5 $\times 10^{3}$ m/s = 346.8 km/s

Therefore, the maximum speed of the emitted photoelectrons is 346.8 km/s.

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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:

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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
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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).

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$_{} \frac{}{}$ $K . E_{1} = \frac{1230}{800} - ?$ - (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 __________.

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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

$f (x) = {\begin{matrix} \frac{2 x^{2} - 3 x + 2}{x - 2}, & x \neq 2 \\ 5, & x = 2 \end{matrix}$

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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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11.2 The work function of cesium metal is 2.14 eV. When light of frequency 6 ×1014 Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) Maximum kinetic energy of the emitted electrons, (b) Stopping potential, and (c) Maximum speed of the emitted photoelectrons?

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