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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

11.26 Ultraviolet light of wavelength 2271 Å from a 100 W mercury source irradiates a photo-cell made of molybdenum metal. If the stopping potential is –1.3 V, estimate the work function of the metal. How would the photo-cell respond to a high intensity ( 10⁵ W m^–2) red light of wavelength 6328 Å produced by a He-Ne laser?

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Payal Gupta | Contributor-Level 10

5 months ago

11.26 Wavelength of ultraviolet light, $λ$ = 2271Å = 2271 $\times 10^{- 10}$ m

Stopping potential of the metal, $V_{0}$ = 1.3 V

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

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

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

Work function of the metal, $\emptyset_{0}$

Frequency of light = $ν$

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

$\emptyset_{0} = h ν - e V_{0}$

= $\frac{h c}{λ} - e V_{0}$

= $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{2271 \times 10^{- 10}} - 1.6 \times 10^{- 19} \times 1.3$

= 8.75 $\times 10^{- 19}$ - 2.08 $\times 10^{- 19}$

= 6.67 $\times 10^{- 19}$ J

= $\frac{6.67 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ eV

= 4.17 eV

Let $ν_{0}$ be the threshold frequency of the metal.

Therefore, $\emptyset_{0}$ = h $ν_{0}$

$ν_{0} =$ $\frac{\emptyset_{0}}{h}$ = $\frac{6.67 \times 10^{- 19}}{6.626 \times 10^{- 34}}$ Hz = 1.007 $\times 10^{15}$ Hz

Wavelength of red light, $λ_{r}$ &nbs

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11.26 Wavelength of ultraviolet light, $λ$ = 2271Å = 2271 $\times 10^{- 10}$ m

Stopping potential of the metal, $V_{0}$ = 1.3 V

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

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

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

Work function of the metal, $\emptyset_{0}$

Frequency of light = $ν$

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

$\emptyset_{0} = h ν - e V_{0}$

= $\frac{h c}{λ} - e V_{0}$

= $\frac{6.626 \times 10^{- 34} \times 3 \times 10^{8}}{2271 \times 10^{- 10}} - 1.6 \times 10^{- 19} \times 1.3$

= 8.75 $\times 10^{- 19}$ - 2.08 $\times 10^{- 19}$

= 6.67 $\times 10^{- 19}$ J

= $\frac{6.67 \times 10^{- 19}}{1.6 \times 10^{- 19}}$ eV

= 4.17 eV

Let $ν_{0}$ be the threshold frequency of the metal.

Therefore, $\emptyset_{0}$ = h $ν_{0}$

$ν_{0} =$ $\frac{\emptyset_{0}}{h}$ = $\frac{6.67 \times 10^{- 19}}{6.626 \times 10^{- 34}}$ Hz = 1.007 $\times 10^{15}$ Hz

Wavelength of red light, $λ_{r}$ = 6328 Å = 6328 $\times 10^{- 10}$ m

Frequency of red light, $ν_{r}$ = $\frac{c}{λ_{r}}$ = $\frac{3 \times 10^{8}}{6328 \times 10^{- 10}}$ = 4.74 $\times 10^{14}$ Hz

Since $ν_{0} > ν_{r}, t h e p h o t o c e l l w i l l n o t r e s p o n d t o t h e r e d l i g h t p r o d u c e d b y t h e l a s e r .$

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hu = hu₀ + K.E

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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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