# 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 .

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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

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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