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

get let p and e denote the linear momentum and energy of a photon from screen.

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

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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.

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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

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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]

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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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