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    let p and e denote the linear momentum and energy of a photon

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    Let ' p ' and ' E ' denote the linear momentum and energy of emitted photon respectively. If the wavelength of incident radiation is increased .

    Click here👆to get an answer to your question ✍️ Let ' p ' and ' E ' denote the linear momentum and energy of emitted photon respectively. If the wavelength of incident radiation is increased .

    Question

    Let 'p' and 'E' denote the linear momentum and energy of emitted photon respectively. If the wavelength of incident radiation is increased ___________.

    A

    Both p and E increase

    B

    p increases and E decreases

    C

    p decreases and E increases

    D

    Both p and E decrease

    Easy Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is D)

    Energy of photon  E=

    λ hc ​ ⟹   E∝ λ 1 ​

    Hence energy of photon decreases as the wavelength of light increases.

    Also, momentum of the photon  p=

    c E ​ ⟹ p∝E

    Thus momentum of photon also decreases as its energy decreases.

    Hence option D is correct.

    Video Explanation

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    Let P and E denote the linear momentum and energy of emitted photon respectively. If the wavelength of incident radiation is increased, then

    hv=W(0)+1/2mv("max")^(2) or (hc)/(lambda)=W(0)+1/2 mv("max")^(2) If lambda is increased, energy of incident radiation is decreased . The work function remains constant . :.1/2 mv^(2) will decrease and momentum p =mv will decrease Thus both p and E will decrease.

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    Let P and E denote the linear ...

    Let P and E denote the linear momentum and energy of emitted photon respectively. If the wavelength of incident radiation is increased, then

    Updated On: 27-06-2022

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    both p and E increase

    B

    p increases and E decreases

    C

    p decreases and E increases

    D

    both p and E decrease

    Answer

    The correct Answer is D

    `hv=W_(0)+1/2mv_("max")^(2)`

    Solution

    or `(hc)/(lambda)=W_(0)+1/2 mv_("max")^(2)` If `lambda` is increased, energy of incident radiation is decreased . The work function remains constant . `:.1/2 mv^(2)` will decrease and momentum p `=`mv will decrease Thus both p and E will decrease.

    Answer

    Step by step solution by experts to help you in doubt clearance & scoring excellent marks in exams.

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    Let P and E Denote the Linear Momentum and Energy of a Photon. If the Wavelength is Decreased,

    Let P and E Denote the Linear Momentum and Energy of a Photon. If the Wavelength is Decreased,

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    Let p and E denote the linear momentum and energy of a photon. If the wavelength is decreased,

    OPTIONS

    both p and E increase

    p increases and E decreases

    p decreases and E increases

    both p and E decrease

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    SOLUTION

    both p and E increase

    From the de-Broglie relation, wavelength,

    λ λ=hp ....(1) ⇒ λ p=hλ

    Here, h = Planck's constant

    p = momentum of electron

    It is clear from the above equation that

    λ p∝1λ .

    Thus, if the wavelength λ

    (λ)

    is decreased, then momentum

    (p) will be increase .

    Relation between momentum and energy :

    p=2mE

    Here, E = energy of electron

    m = mass of electron

    Substituting the value of p in equation (1), we get :

    λ λ=h2mE ⇒ λ E=hλ2m ⇒ λ E=h22mλ2

    Thus, on decreasing λ , the energy will increase .

    Concept: de-Broglie Relation

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    Chapter 20: Photoelectric Effect and Wave-Particle Duality - MCQ [Page 363]

    Q 3 Q 2 Q 4

    APPEARS IN

    HC Verma Class 11, Class 12 Concepts of Physics Vol. 2

    Chapter 20 Photoelectric Effect and Wave-Particle Duality

    MCQ | Q 3 | Page 363

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