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In an excited state of hydrogen like atom an electron has total energy of `-3.4 eV`. If the kinetic energy of the electron is E and its de-Broglie wavelength is `lambda`, thenA. `E = 6.8 eV: lambda = 6.6 xx 10^(-10) m`B. `E = 3.4 eV: lambda = 6.6 xx 10^(-10) m`C. `E = 3.4 eV: lambda = 6.6 xx 10^(-11) m`D. `E = 6.8 eV: lambda = 6.6 xx 10^(-11) m` |
Answer» Correct Answer - B Potential energy `= - 2 xx` kinetic energy = `- 2 E` Total energy `= - 2 E + E = - 3.4 eV = - E` `E = 3.4 eV` `P = "momentum"`, m = mass of electron or `P = sqrt(2 m E)` `= sqrt(2 xx 9.1 xx 10^(-31) xx 3.4 xx 1.6 xx 10^(-19)) = 10^(-24)` de Brogile wavelength `lambda = (h)/(P) = (h)/(sqrt(2 m E)) = 6.6 xx 10^(-10) m` |
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