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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

11.27 Monochromatic radiation of wavelength 640.2 nm (1nm = 10^–9 m) from a neon lamp irradiates photosensitive material made of cesium on tungsten. The stopping voltage is measured to be 0.54 V. The source is replaced by an iron source and its 427.2 nm line irradiates the same photo-cell. Predict the new stopping voltage.

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Payal Gupta | Contributor-Level 10

4 months ago

11.27 Wavelength of the monochromatic light, $λ$ = 640.2 nm = 640.2 ${\times 10}^{- 9} m$

Stopping potential of neon lamp, $V_{0}$ = 0.54 V

Charge of an electron, e = 1.6 $\times 10^{- 19} C$

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

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

Let $\emptyset_{0}$ be the work function and $ν$ frequency of emitted light

We have the photo-energy relation from the photoelectric effect as:

$e V_{0}$ = $h ν - \emptyset_{0}$ = h $\frac{c}{λ}$ - $\emptyset_{0}$

$\emptyset_{0} = \frac{h c}{λ}$ - $e V_{0}$

= $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{640.2 {\times 10}^{- 9}}$ $-$ 1.6 $\times 10^{- 19} \times 0.54$

= 3.105 $\times 10^{- 19}$ - 0.864 $\times 10^{- 19}$

= 2.241 $\times 10^{- 19}$ J

= $\frac{2.241 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ eV

=1.40 eV

The wavelength of the radiation emitted from an iron source, $λ'$ = 427.2 nm = 427.2

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11.27 Wavelength of the monochromatic light, $λ$ = 640.2 nm = 640.2 ${\times 10}^{- 9} m$

Stopping potential of neon lamp, $V_{0}$ = 0.54 V

Charge of an electron, e = 1.6 $\times 10^{- 19} C$

Planck’s constant, h = 6.626 $\times 10^{- 34}$ Js

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

Let $\emptyset_{0}$ be the work function and $ν$ frequency of emitted light

We have the photo-energy relation from the photoelectric effect as:

$e V_{0}$ = $h ν - \emptyset_{0}$ = h $\frac{c}{λ}$ - $\emptyset_{0}$

$\emptyset_{0} = \frac{h c}{λ}$ - $e V_{0}$

= $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{640.2 {\times 10}^{- 9}}$ $-$ 1.6 $\times 10^{- 19} \times 0.54$

= 3.105 $\times 10^{- 19}$ - 0.864 $\times 10^{- 19}$

= 2.241 $\times 10^{- 19}$ J

= $\frac{2.241 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ eV

=1.40 eV

The wavelength of the radiation emitted from an iron source, $λ'$ = 427.2 nm = 427.2 $\times 10^{- 9}$ m

Let the new stopping potential be $V_{0}'$

From the relation $\emptyset_{0} = \frac{h c}{λ'}$ - $e V_{0}'$ , we get

$e V_{0}^{'} = \frac{h c}{λ^{'}} - \emptyset_{0}$

= $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{427.2 \times 10^{- 9}}$ - 2.241 $\times 10^{- 19}$

= 4.653 $\times 10^{- 19}$ - 2.241 $\times 10^{- 19}$

= 2.412 $\times 10^{- 19}$ J

= $\frac{2.412 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ eV

= 1.51 eV

The new stopping potential is 1.51 eV.

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$\begin{array}{l} f (x) = \frac{2 x^{2} - 3 x - 2}{x - 2} \\ = \frac{2 x^{2} - 4 x + x - 2}{x - 2} = \frac{2 x (x - 2) + 1 (x - 2)}{x - 2} \\ = \frac{(2 x + 1) (x - 2)}{x - 2} = 2 x + 1 \\ \underset{x \to 2^{-}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 - h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2^{+}}{l i m} f (x) = 2 x + 1 = \underset{h \to 0}{l i m} 2 (2 + h) + 1 = 4 + 1 = 5 \\ \underset{x \to 2}{l i m} f (x) = 5 \\ A s \underset{x \to 2^{-}}{l i m} f (x) = \underset{x \to 2^{+}}{l i m} f (x) = \underset{x \to 2}{l i m} f (x) = 5 \\ H e n c e, f (x) i s c o n t i n u o u s a t x = 2 . \end{array}$

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