The wavelength lamda_(e) of ann electron and lamda_(p) of a photon of same energy E are related by
The wavelength lambda_(e ) of an photon of same energy E are related by
Momentum of a photon of wavelength lambda is :
According to de-Broglie hypothesis, the wavelength associated with moving electron of mass 'm' is 'lambda_(e)' . Using mass energy relation and Planck's quantum theory, the wavelength associated with photon is 'lambda_(p)' . If the energy (E) of electron and photonm is same, then relation between lambda_e and 'lambda_(p)' is
The ratio of the wavelengths of a photon and that of an electron of same energy E will be [m is mass of electron ]
If the wavelength lambda of photon decreases then momentum and energy of photon
The debroglie wavelength of an electron is lambda and has energy E. If the debroglie wavelength is reduced by 25% then how much extra kinetic energy will the electron have now?
Neglecting the mass variation with velocity , the ratio of the wavelength ((lamda_e)/lamda_p) associated with an electron (lamda_e) having a kinetic energy E and wavelength associated with a Photon (lamda_p) having kinetie energy 4E is
Photon and electron are given same energy (10^(-20) J) . Wavelength associated with photon and electron are lambda_(ph) and lambda_(el) then correct statement will be
AAKASH INSTITUTE-ALTERNATING CURRENT -Assignment (Section-J) (Aakash Chailengers Questions)
