In the decay `^64(Cu) rarr `^64(Ni)+e^+v,`the maximum kinetic energy carried by the netrino which was emitted together with a positron of kinetic energy 0.150 meV? (b)What is the momentum of this neutrino in kg m `s^(-1)`?Use the formula applicable to photon.

In the decay Cu^64 rarr Ni^64 + e^+ + v , the maximum kinetic energy carried by the netrino which was emitted together with a positron of kinetic energy 0.150 meV? (a) what is the energy of the neutrino which was emitted togather with a positron of kinetic energy 0.150 MeV ? .(b)What is the momentum of this neutrino in kg m s^(-1) ?Use the formula applicable to photon.

What is the momentum of a photon of energy 1 MeV in kg m/s ?

An object is moving so that its kinetic energy is 150 J and the magnitude of its momentum is 30.0 kg-m/s. With what velocity is it travelling?

An object is moving so that its kinetic energy is 150 J and the magnitude of its momentum is 30.0 kg-m/s. With what velocity is it travelling?

An object is moving so that its kinetic energy is 150 J and the magnitude of its momentum is 30.0 kg-m/s. With what velocity is it travelling?

Gold ._(79)^(198)Au undergoes beta^(-) decay to an excited state of ._(80)^(198)Hg . If the excited state decays by emission of a gamma -photon with energy 0.412 MeV , the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligble energy. The recoil enregy of the ._(80)^(198)Hg nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968255 u for ._(79)^(198)Hg ).

Gold ._(79)^(198)Au undergoes beta^(-) decay to an excited state of ._(80)^(198)Hg . If the excited state decays by emission of a gamma -photon with energy 0.412 MeV , the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligble energy. The recoil enregy of the ._(80)^(198)Hg nucleus can be ignored. The masses of the neutral atoms in their ground states are 197.968255 u for ._(79)^(198)Hg ).

The work fuction of caseium is 1.8 eV. Light of 4500Å is incident on it. Calculate (i) the maximum kinetic energy of the emitted photoelectrons (ii) maximum velocity of the emitted photoelectrons (iii) if the intensity of the incident light is doubled, then find the maximum kinetic energy of the emitted photoelectrons. Given, h=6.63xx10^(-34)Js.m_(e)=9.1xx10^(-31)kg, c=3xx10^(8)ms^(-1)