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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

11.28 A mercury lamp is a convenient source for studying frequency dependence of photoelectric emission, since it gives a number of spectral lines ranging from the UV to the red end of the visible spectrum. In our experiment with rubidium photo-cell, the following lines from a mercury source were used:

$λ_{1}$ = 3650 Å, $λ_{2}$ = 4047 Å, $λ_{3}$ = 4358 Å, $λ_{4}$ = 5461 Å, $λ_{5}$ = 6907 Å,

The stopping voltages, respectively, were measured to be:

$V_{01}$ = 1.28 V, $V_{02}$ = 0.95 V, $V_{03}$ = 0.74 V, $V_{04}$ = 0.16 V, $V_{05}$ = 0 V

Determine the value of Planck’s constant h, the threshold frequency and work function for the material.

[Note: You will notice that to get h from the data, you will need to know e (which you can take to be 1.6 $\times 10^{- 19}$ C). Experiments of this kind on Na, Li, K, etc. were performed by Millikan, who, using his own value of e (from the oil-drop experiment) confirmed Einstein’s photoelectric equation and at the same time gave an independent estimate of the value of h.]

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Payal Gupta | Contributor-Level 10

3 months ago

11.28 Einstein’s photoelectric equation is given as:

$e V_{0}$ = $h ν - \emptyset_{0}$

$V_{0} = \frac{h}{e} ν - \frac{\emptyset_{0}}{e}$ ……….(1)

where

$V_{0} =$ Stopping potential

h = Planck’s constant

e = Charge of an electron

$ν = F r e q u e n c y o f r a d i a t i o n$

$\emptyset_{0} = W o r k f u n c t i o n o f a m a t e r i a l$

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

It can be concluded from equation (1) that $V_{0}$ is directly proportional to frequency $ν$

Now frequency $ν$ can be expressed as

$ν = \frac{c}{λ}$

Then,

$ν_{1}$ = $\frac{c}{λ_{1}}$ = $\frac{3 \times 10^{8}}{3650 Å}$ Hz = 8.219 $\times 10^{14}$ Hz

$ν_{2}$ = $\frac{c}{λ_{2}}$ = $\frac{3 \times 10^{8}}{4047 Å}$ = $\frac{3 \times 10^{8}}{3650 \times 10^{- 10}}$ Hz = 7.413 $\times 10^{14}$
Hz

$ν_{3}$ = $\frac{c}{λ_{3}}$ = $\frac{3 \times 10^{8}}{4358 Å}$ = $\frac{3 \times 10^{8}}{4047 \times 10^{- 10}}$ Hz = 6.883 $\times 10^{14}$ Hz

$ν_{4}$ = $\frac{c}{λ_{4}}$ = $\frac{3 \times 10^{8}}{5461 Å}$ = $\frac{3 \times 10^{8}}{4358 \times 10^{- 10}}$ Hz = 5.493 $\times 10^{14}$ Hz

$ν_{5}$ = $\frac{c}{λ_{5}}$ = $\frac{3 \times 10^{8}}{6907 Å}$ = $\frac{3 \times 10^{8}}{5461 \times 10^{- 10}}$ Hz = 4.343 $\times 10^{14}$ Hz

From the given data of

...more

11.28 Einstein’s photoelectric equation is given as:

$e V_{0}$ = $h ν - \emptyset_{0}$

$V_{0} = \frac{h}{e} ν - \frac{\emptyset_{0}}{e}$ ……….(1)

where

$V_{0} =$ Stopping potential

h = Planck’s constant

e = Charge of an electron

$ν = F r e q u e n c y o f r a d i a t i o n$

$\emptyset_{0} = W o r k f u n c t i o n o f a m a t e r i a l$

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

It can be concluded from equation (1) that $V_{0}$ is directly proportional to frequency $ν$

Now frequency $ν$ can be expressed as

$ν = \frac{c}{λ}$

Then,

$ν_{1}$ = $\frac{c}{λ_{1}}$ = $\frac{3 \times 10^{8}}{3650 Å}$ Hz = 8.219 $\times 10^{14}$ Hz

$ν_{2}$ = $\frac{c}{λ_{2}}$ = $\frac{3 \times 10^{8}}{4047 Å}$ = $\frac{3 \times 10^{8}}{3650 \times 10^{- 10}}$ Hz = 7.413 $\times 10^{14}$
Hz

$ν_{3}$ = $\frac{c}{λ_{3}}$ = $\frac{3 \times 10^{8}}{4358 Å}$ = $\frac{3 \times 10^{8}}{4047 \times 10^{- 10}}$ Hz = 6.883 $\times 10^{14}$ Hz

$ν_{4}$ = $\frac{c}{λ_{4}}$ = $\frac{3 \times 10^{8}}{5461 Å}$ = $\frac{3 \times 10^{8}}{4358 \times 10^{- 10}}$ Hz = 5.493 $\times 10^{14}$ Hz

$ν_{5}$ = $\frac{c}{λ_{5}}$ = $\frac{3 \times 10^{8}}{6907 Å}$ = $\frac{3 \times 10^{8}}{5461 \times 10^{- 10}}$ Hz = 4.343 $\times 10^{14}$ Hz

From the given data of stopping potential, we get the following table

Frequency $\times 10^{14}$ Hz	8.219	7.413	6.883	5.493	4.343
Stopping potential $V_{0}$ V	1.28	0.95	0.74	0.16	0

It can be observed that the obtained curve is a straight line. It intersects the frequency axis at 5 $\times 10^{14}$ Hz, which is the threshold frequency ( $ν_{0}$
) of the material. Point D corresponds to a frequency less than threshold frequency. Hence, there is no photoelectric emission for the $λ_{5}$ line and therefore, no stopping voltage is required to stop the current.

Slope of the straight line = $\frac{A B}{B C}$ = $\frac{1.28 - 0.16}{(8.219 - 5.493) \times 10^{14}}$

From equation (1), the slope $\frac{h}{e}$
can be written as

$\frac{h}{e} = \frac{1.28 - 0.16}{(8.219 - 5.493) \times 10^{14}}$ = $\frac{1.12}{2.726 \times 10^{14}}$ Js

$h$ = $\frac{1.12}{2.726 \times 10^{14}} \times 1.6 \times 10^{- 19}$ = 6.573 $\times 10^{- 34}$ Js

The work function of the material is given by, $\emptyset_{0}$ = $h ν_{0}$ = 6.573 $\times 10^{- 34} \times$ 5 $\times 10^{14}$

= 3.286 $\times 10^{- 19}$ J = $\frac{3.286 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ eV = 2.054eV

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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:

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)
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h/√2mk? = (1/2.3) × h/√2mk?
=> k? = (2.3)²k? = 5.25k? (ii)
Now, k? = E? - 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)

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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)

$\frac{v_{2}}{v_{1}} = \sqrt[]{\frac{8}{2} = 2}$

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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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