The ratio isotope `._(29)^(64)Cu` can decay through positron emission and electron capture. In both the processes neutrino (v) is emitted. Calculate the difference in maximum possible kinetic energy of emitted neutrinos in the two processess.
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In photoelectric emission, the energy of the emitted electron is
Correct Answer - C (c) In the photoelectric effect, the energy of the emitted electron is smaller than that of the incident photon because some energy of photon is used to...
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If `X^(a)` species emit firstly a positron then two `alpha` and `beta` last one `alpha` is also emitted and finally convert in `Y^(c )` species so cor
Correct Answer - A `_(92)U^(235)` nucleus absorbs a neutron and then disintegrate in `_(54)Xe^(139),38Sr^(94) and X.` Thus `_(92)U^(235)+_(0)n^(1)rarr_(54)Xe^(139)+_(38)Sr^(94)+3_(0)n^(1)`
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`.^(40)K` is an unusual isotope, in that it decays by negative beta emission, positive beta emission, and electron capture. Find the Q values for thes
The process for negative beta decay is given by equation `._(19)^(40)K_(21) rarr ._(20)^(40)Ca_(20) +e^(-) + barv` and the Q value is found from equation using atomic masses : `Q_(beta) [m(.^(40)K)...
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In an electron capture process, a nucleus captures an electron from K shell of the atom. The electron is having a binding energy of `B_(0)`. Followed
Correct Answer - `B_(0)`
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Nuclei of `.^(64)Cu` can decay be electron capture (probability 61%) or by `beta^(+)` decay (39%). Half life of `.^(64)Cu` is 12.7 hour. Find the part
Correct Answer - 20.8hr.
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The energy of K-electron in tungsten is -20.0 keV and of an L-electron is -2.0 ke V. Find the wavelength of X-rays emitted when an electron jumps from
Correct Answer - 0.6875 Å
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Potassium-40 can decay in three modes .It can decay by `beta^(-)` - emission,`beta^(+)` -emission or electron capature. (a) Write the equation showing
Correct Answer - [(a) `._(19)^(40)K rarr ._(20)^(40)Ca + e^(-) + bar(v), ._(19)^(40)K rarr ._(18)^(40)Ar + e^(+) +v, ._(19)^(40)K + e^(-1) rarr ._(18)^(40)Ar+v` (b) 1.3034 MeV, 0.4676 MeV, 1.490 MeV]
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Find the minimum energy that a `gamma`-ray must have to give rise to an electron -positron pair. `{:("Mass of electron",= 0.00055 am u,),("Mass of pos
Correct Answer - 1.024 MeV
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Calculate the amount of energy set free by the annigilation of an electron and a positron. Given that mass of electron `= 0.00055` amu and positron =
Correct Answer - 1.0211 MeV
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Match the Column-I with Column-II: `{:("Column-I","Column-II"),((a) n to p^(+) + ......,(p) "Positron emission"),((b) p^(+) to n + .....,(q) beta - "e
Correct Answer - (a-q)(b-p)(c-r)(d-s)
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