Under what potential difference should an electron be accelerated to obtain de Broglie, wavelength of `0.6overset@A`? `(h=6.62xx10^(-34)j.s, m_e=9.1xx10^(-31)`kg)

Under what potential difference should an electron be accelerated to obtain electron waves of lamda = 0.6 Å ? Given, mass of electron, m = 9.1 xx 10^(-31) kg , Planck's constant, h = 6.62 xx 10^(-34) J.s

What should be the minimum value of potential difference to be applied across a Coolidge tube for production of X-rays of wavelength 0.8 Å ? h=6.62xx10^(-34)J*s,e=1.6xx10^(-19) C

Calculate the velocity of an electron having de Broglie wavelength of 200 Å. [m=9.11xx10^(-31)kg,h=6.626xx10^(-34)J*s] .

Find the wavelength of an electron having kinetic energy 10 eV. (h = 6.33 xx 10^(-34) J.s, m_(e) = 9 xx 10^(-31) kg)

A moving electron has 4.9 xx 10^(-25) joules of kinetic energy. Find out its de - Broglie wavelength (Given h = 6.626 xx 10^(-34) Js, m_e = 9.1 xx 10^(-31) kg).

Calculate the ratio of velocities of a moving electron to that of a proton associated with the same de Broglie wavelength. [m_(e)=9.11xx10^(-31)kg,m_(p)=1.67xx10^(-27)kg,h=6.626xx10^(-34)J*s ]

Calculate the momentum of a particle which has a de-Broglie wavelength of 1Å. [ h = 6.626 xx 10^(-34) kg m^2 s^(-1) ]

Einstein established the idea of photons on the basis of Planck's quantum theory. According to his idea, the light of frequency f or wavelength lamda is infact a stream of photons. The rest mass of each photon is zero and velocity is equal to the velocity of light (c) = 3 xx 10^(8) m.s^(-1) . Energy, E = hf, where h = Planck's constant = 6.625 xx 10^(-34)J.s . Each photon has a momentum p = (hf)/(c) , although its rest mass is zero. The number of photons increase when the intensity of incident light increases and vice-versa. On the other hand, according to de Broglie any stream of moving particles may be represented by progressive waves. The wavelength of the wave (de Broglie wavelength) is lamda = (h)/(p) , where p is the momentum of the particle. When a particle having charge e is accelerated with a potential difference of V, the kinetic energy gained by the particle is K= eV. Thus as the applied potential difference is increased, the kinetic energy of the particle and hence the momentum increase resulting in a decrease in the de Broglie wavelength. Given, charge of electron, e = 1.6 xx 10^(-19)C and mass = 9.1 xx 10^(-31) kg . The number of photons emitted per second from a light source of power 40 W and wavelength 5893 Å

If the mass of neutron is 1.7xx10^(-27) kg , then the de Broglie wavelength of neutron of energy 3eV is (h = 6.6xx10^(-34) J.s)