Two monochromatic beams A and B of equal intensity I, hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then, what inference can you make about their frequencies?

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    Answered by

    alok kumar singh | Contributor-Level 10

    3 months ago

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

    Explanation- Suppose nA is the number of photons falling per second of beam A and nB is the number of photons falling per second of beam B.

    NA=2nB

    energy of falling photon A=hvA

    B=hvB

    as we know intensities are same

    nAhvA=nBhvB

    va/vb=nB/nA=1/2

    vB=2vA

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Raj Pandey

Based on theory

V
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
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V
Vishal Baghel

  K . E 1 = 1 2 3 0 8 0 0 ? - (i)

  K . E 2 = 2 K . E 1 = 1 2 3 0 5 0 0 ? - (ii)

from (i) & (ii)

? = 0 . 6 1 5  ev

V
Vishal Baghel

hu = hu0 + K.E

Cases u = 2u0

h2u0 = hu0 + K.E1

K.E1 = hu0

1 2 m v 1 2 = h υ 0

v 1 2 = 2 h υ 0 m  

v 1 = 2 h υ 0 m - (1)      

Now, cases 2

h 5u0 = hu0 + k.E2

k.E2 = 4hu0

1 2 m v 2 2 = 4 h υ 0

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A
alok kumar singh

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

Sol:

f(x)=2x23x2x2=2x24x+x2x2=2x(x2)+1(x2)x2=(2x+1)(x2)x2=2x+1limx2f(x)=2x+1=limh02(2h)+1=4+1=5limx2+f(x)=2x+1=limh02(2+h)+1=4+1=5limx2f(x)=5Aslimx2f(x)=limx2+f(x)=limx2f(x)=5Hence,f(x)iscontinuousatx=2.

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