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Dual Nature of Radiation and Matter Physics Ncert Solutions Class 12th Physics Class 12th

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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Vishal Baghel | Contributor-Level 10

3 months ago

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

hu = hu0 + K.ECases u = 2u0h2u0 = hu0 + K.E1K.E1 = hu0 <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>m</mi> <msubsup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>=</mo> <mi>h</mi> <msub> <mrow> <mi>υ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> </math> <math> <mrow> <msubsup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>=</mo> <mfrac> <mrow> <mn>2</mn> <mi>h</mi> <msub> <mrow> <mi>υ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> </mrow> </math>   <math> <mrow> <msub> <mrow> <mi>v</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> <mo>=</mo> <mroot> <mrow> <mfrac> <mrow> <mn>2</mn> <mi>h</mi> <msub> <mrow> <mi>υ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math> - (1)      Now, cases 2h 5u0 = hu0 + k.E2k.E2 = 4hu0 <math> <mrow> <mfrac> <mrow> <mn>1</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mi>m</mi> <msubsup> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> <mrow> <mn>2</mn> </mrow> </msubsup> <mo>=</mo> <mn>4</mn> <mi>h</mi> <msub> <mrow> <mi>υ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> </math> v2 = <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1029" DrawAspect="Content" ObjectID="_1814702160"> </o:OLEObject></xml><! [endif]->    <math> <mrow> <mroot> <mrow> <mfrac> <mrow> <mn>8</mn> <mi>h</mi> <msub> <mrow> <mi>υ</mi> </mrow> <mrow> <mn>0</mn> </mrow> </msub> </mrow> <mrow> <mi>m</mi> </mrow> </mfrac> </mrow> <mrow></mrow> </mroot> </mrow> </math>   - (2) <math> <mrow> <mfrac> <mrow> <msub> <mrow> <mi>v</mi> </mrow> <mrow> <mn>2</mn> </mrow> </msub> </mrow> <mrow> <msub> <mrow> <mi>v</mi> </mrow> <mrow> <mn>1</mn> </mrow> </msub> </mrow> </mfrac> <mo>=</mo> <mroot> <mrow> <mfrac> <mrow> <mn>8</mn> </mrow> <mrow> <mn>2</mn> </mrow> </mfrac> <mo>=</mo> <mn>2</mn> </mrow> <mrow></mrow> </mroot> </mrow> </math>        <!- [if gte mso 9]><xml> <o:OLEObject Type="Embed" ProgID="Equation.DSMT4" ShapeID="_x0000_i1030" DrawAspect="Content" ObjectID="_1814702161"> </o:OLEObject></xml><! [endif]-> v2 = 2v1

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Similar Questions for you

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]

Vishal Baghel

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
∴ k? /k? = (10.2 - Φ)/ (4 × 10.2 - Φ) = 1/5.25
=> Φ = 3eV

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).

Vishal Baghel

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

$_{}_{} \frac{}{}$ $K . E_{2} = 2 K . E_{1} = \frac{1230}{500} - ?$ - (ii)

from (i) & (ii)

$? = 0.615$ ev

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:

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

There are two sources of light, each emitting with a power of 100 W. One emits X-rays of wavelength 1 nm and the other visible light at 500 nm. Find the ratio of number of photons of X-rays to the photons of visible light of the given wavelength.

alok kumar singh

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

Explanation- p=100W

_{$λ$
1}= 1nm

_{$λ$
2}= 500nm

Also n₁and n₂ represent number of photon of x-rays and visible light emitted from the two sources per sec

E/t=P=n₁hc/ $λ$ ₁₌n₂hc/ $λ$ ₂

So n₁/ $λ$ ₁= n₂/ $λ$ ₂

So n₁/n₂= 1/500

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