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The electron in a given Bohr orbit has a total energy of \( -1.5 eV \). Calculate its (i) Kinetic energy. (ii) potential energy. (iii) Wavelength of radiation emitted, when this electron makes a transition to the ground state. [Given : Energy in the ground state \( =-13.6 eV \) and Rydberg's constant \( =1.09 \times 10^{7} m ^{1} \) ] |
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Answer» Given E = 1.5 ev (i) Kinetic energy of electron K.E. = -E K.E. = -5(-1.5 ev) K. E. = 1.5 ev (ii) Potential energy of electron P.E. = -2 K.E. P.E. = -2 x 1.5 P.E. = -3.0 ev (iii) Energy of photon \(\Delta\)E = E2 - E1 \(\Delta\)E = -1.5 - (-13.6) \(\Delta\)E = 12.1 ev \(\Delta\)E = 12.1 x 1.6 x 10-19 J \(\Delta\)E = 19.36 x 10-19 J Energy of photon \(\Delta\)E = \(\frac{hc}\lambda\) \(\frac{hc}\lambda\) = 19.36 x 10-19 Wavelength of radiation \(\lambda\) = \(\frac{6.6\times10^{-34}\times3\times10^8}{19.36\times10^{-19}}\) \(\lambda=1.022\times10^{-7}\) m |
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