Nuclear Physics and RadioactivityHard
Question
An electron is in an excited stable in hydrogen - like atom. It has a total energy of -3.4 eV. If the kinetic energy of the electron is E and its de-Broglie wavelength is λ,then
Options
A.E = 6.8 eV, λ = 6.6 × 10-10 m
B.E = 3.4 eV, λ = 6.6 × 10-10 m
C.E = 3.4 eV, λ = 6.6 × 10-11 m
D.E = 6.8 eV, λ = 6.6 × 10-11 m
Solution
P.E. = - 2(K.E.)
T.E. = (P.E.) + (K.E.)
T.E = - 2(K.E.) + (K.E.)
T.E. = - (K.E.) - T.E. = K.E. K.E. = 3.4 eV
λ = 6.6 × 10-10 m
T.E. = (P.E.) + (K.E.)
T.E = - 2(K.E.) + (K.E.)
T.E. = - (K.E.) - T.E. = K.E. K.E. = 3.4 eV
λ = 6.6 × 10-10 m
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