11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 <math><mo>×</mo><mi mathvariant="bold-italic"></mi><msup><mrow><mrow><mn>10</mn></mrow></mrow><mrow><mrow><mo>-</mo><mn>15</mn></mrow></mrow></msup></math> Vs. Calculate the value of Planck’s constant.

11. The slope of the cut-off voltage (V) versus frequency ( ν) of an incident light is given as:Vν= 4.12 ×10-15 VsThe relationship of V and ν is given as hν = eV or Vν= hewhere e = Charge of an electron = 1.6 ×10-19 C and h = Plank’s constantTherefore, h = e×Vν = 1.6 ×10-19 × 4.12 ×10-15 = 6.592 ×10-34 JsPlank’s constant = 6.592 ×10-34 Js

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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

11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 $\times 10^{- 15}$ Vs. Calculate the value of Planck’s constant.

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Payal Gupta | Contributor-Level 10

4 months ago

11. The slope of the cut-off voltage (V) versus frequency ( $ν)$ of an incident light is given as:

$\frac{V}{ν} =$ 4.12 $\times 10^{- 15}$ Vs

The relationship of V and $ν$ is given as $h ν$ = $e V$ or $\frac{V}{ν} =$ $\frac{h}{e}$

where e = Charge of an electron = 1.6 $\times 10^{- 19}$ C and h = Plank’s constant

Therefore, h = $e \times \frac{V}{ν}$ = 1.6 $\times 10^{- 19}$ $\times$ 4.12 $\times 10^{- 15}$ = 6.592 $\times 10^{- 34}$ Js

Plank’s constant = 6.592 $\times 10^{- 34}$ Js

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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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hu = hu₀ + K.E

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This is a Short Answer Type Questions as classified in NCERT Exemplar

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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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11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 $\times 10^{- 15}$ Vs. Calculate the value of Planck’s constant.

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11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 ×10-15 Vs. Calculate the value of Planck’s constant.

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11.6 In an experiment on photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 $\times 10^{- 15}$ Vs. Calculate the value of Planck’s constant.