Q5 of 805 Page 1

Two particles have equal momenta. What is the ratio of their de-Broglie wavelengths?

OR


Monochromatic light of frequency 6.0x 1014 Hz is produced by a laser. What is the energy of a photon in the light beam?

Given: -


Two particles have equal momenta ‘p’


Formula: -


the de Broglie wavelength associated with a given particle is,



Where λ is the de-Broglie wavelength of the particle, h is the Planck’s constant and p is the momentum.


So, we can write the ratio as



(p1 = p2 = p)


Conclusion: -


The ratio of de-Broglie wavelengths is 1.


OR


Given: -


Frequency of Monochromatic light, v = 6.0x 1014 Hz


Formula: -


We know that the energy of a photon is given by the equation,


E = h v


Where, E is the energy of the photon,


h is the Planck’s constant and,


v is the frequency.


So, substituting the values we get,


E = 6.626 × 10-34 × 6 × 1014= 39.756 × 10-20 J.


Conclusion: -


The energy of the light is, E = 3.9756 × 10-19 J.


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3

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OR


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4

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6

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OR


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7

The diagram below shows a potentiometer set up. On touching the jockey near to the end X of the potentiometer wire, the galvanometer pointer deflects to left. On touching the jockey near to end Y of the potentiometer, the galvanometer pointer again deflects to left but now by a larger amount. Identify the fault in the circuit and explain, using appropriate equations or otherwise, how it leads to such a one-sided deflection.


OR


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(a)Write the formula to be used for finding X from the observations.


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