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calculate de brogile wavelength(in cm) of an α -particle emitted from radium having an energy of 4.8MeV. Mass of α -particle=6.6x10^-24g calculate de brogile wavelength(in cm) of an α-particle emitted from radium having an energy of 4.8MeV. Mass of α-particle=6.6x10^-24g
calculate de brogile wavelength(in cm) of an α-particle emitted from radium having an energy of 4.8MeV. Mass of α-particle=6.6x10^-24g
Dear Prince Solve using the formula: The derivation of De Brogiles wave length: We know that the energy of light E=mc^2 Again if the frequency of light is f Then E=h*f [where h is planck constant] So, mc^2=h*f or, mc^2=h*(c/l) [where c is the velocity of light,l is the wavelength ,& c=l*f] or,l=(h*c)/mc^2 or,l=h/m*c or,l=h/p.[ where p is the momentum=m*c] this is the De Broglies equation. That derivation is completely wrong. The momentum of a photon is NOT given by m*c, as c=constant, m=0. Neither is the energy of light mc^2, as that also goes to zero. The derivation works by a miracle. The proper derivation is this: From a relativistic invariant we know that E^2-p^2c^2=(mc^2)^2 For a photon, the mass = 0, so E/c=p For a photon, E=hc/lambda Therefore, h/lambda=p The first step works as a result of contra/covariant vectors, but can be checked using algebra. The relativistic Energy = m*c^2/sqrt(1-v^2/c^2) , momentum p = m*v/sqrt(1-v^2/c^2). Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple to download the toolbar…. So start the brain storming…. become a leader with Elite Expert League ASKIITIANS Thanks Aman Bansal Askiitian Expert
Dear Prince
Solve using the formula:
The derivation of De Brogiles wave length: We know that the energy of light E=mc^2 Again if the frequency of light is f Then E=h*f [where h is planck constant] So, mc^2=h*f or, mc^2=h*(c/l) [where c is the velocity of light,l is the wavelength ,& c=l*f] or,l=(h*c)/mc^2 or,l=h/m*c or,l=h/p.[ where p is the momentum=m*c] this is the De Broglies equation. That derivation is completely wrong. The momentum of a photon is NOT given by m*c, as c=constant, m=0. Neither is the energy of light mc^2, as that also goes to zero. The derivation works by a miracle. The proper derivation is this: From a relativistic invariant we know that E^2-p^2c^2=(mc^2)^2 For a photon, the mass = 0, so E/c=p For a photon, E=hc/lambda Therefore, h/lambda=p The first step works as a result of contra/covariant vectors, but can be checked using algebra. The relativistic Energy = m*c^2/sqrt(1-v^2/c^2) , momentum p = m*v/sqrt(1-v^2/c^2).
Cracking IIT just got more exciting,It s not just all about getting assistance from IITians, alongside Target Achievement and Rewards play an important role. ASKIITIANS has it all for you, wherein you get assistance only from IITians for your preparation and win by answering queries in the discussion forums. Reward points 5 + 15 for all those who upload their pic and download the ASKIITIANS Toolbar, just a simple to download the toolbar….
So start the brain storming…. become a leader with Elite Expert League ASKIITIANS
Thanks
Aman Bansal
Askiitian Expert
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