The voltage applied across the cathode and anode of an X-ray generating machine is 50000 V. Determine the shortest wavelength of the X-ray emitted. Given ` h = 6.62 xx 10^(-34)" " Jcdot s`.
Or, Write down the equation of `beta`-decay. Why is the detection of neutrinos difficult?

The voltage applied across the cathode and anode of an X-ray generating machine is 50000V. Determine the shortest wavelength of the X-ray emitted. Given h=6.62xx10^(-34)J*s .

The voltage applied across the cathode and anode of an X-ray generating machine is 50,000 V. Determine the shortest wavelength of the X-ray emitted. Given h=6.62xx10^-34 js.

How many photons will a source of power 50W produce per second? The wavelength of an emitted photon is 6000 Å . Given , h = 6.62xx10^(-34)J.s

Calculate the minimum wavelength of x-ray produced by on X-ray tube operating at 30KV [h = 6.6 xx 10^-34Js] .

Estimate the kinetic energy of electron that hit the target in a Coolidge tube and produce X-ray of wavelength 1 Å . Given h=6.62xx10^(-34)J *s,c=3xx10^(8)m*s^(-1) .

If 50 kV potential difference is applied across a Coolidge tube, what will be the maximum frequency of X-ray emitted from the tube? h=6.6xx10^(-34)J*s,e=1.6xx10^(-19)C .

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

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 . Two stationary electrons are accelerated with potential difference V_(1) and V_(2) respectively such that V_(1) : V_(2) = n . The ratio of their de Broglie wavelength

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 by a source of light of power 30 W is 10^(20) , the momentum of each photon (in kg.m.s^(-1) )