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an electron is a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum required for its escape from the atom.what is the wavelength of emitted electron? A)4.70 ANGSTROM B)4.70 NANOMETER C)9.4 ANGSTROM D)9.40 NANOMETER an electron is a hydrogen atom in its ground state absorbs 1.50 times as much energy as the minimum required for its escape from the atom.what is the wavelength of emitted electron? A)4.70 ANGSTROM B)4.70 NANOMETER C)9.4 ANGSTROM D)9.40 NANOMETER
Hi Deepankar, Energy required for the electron to escape is 13.6 eV Excess energy that electron have is 0.5*13.6=6.8 eV 6.8 * 1.6 *10^-19 J = h c/wavelength but the answers are not matching... there might may be some mistake... check it out
excess energy = 6.8 eV = 6.8 * 1.6 * 10^-19 J = 1.09 * 10^-18 Jh/mc = wavelengthhere it can also be written as wavelength = h/sqrt(2mKE)where h = 6.626 * 10^-34m = 9.1 * 10^-31 kgKE = eV = 1.09 * 10^-18 Jso substitute these answers and the right answer is 4.7 angstrom.
Energy required for the electron to escape is 13.6 eV Excess energy that electron have is 0.5*13.6=6.8 eV 6.8 * 1.6 *10^-19 J = h c/wavelength but the answers are not matching... there might may be some mistake... check it out
Excess energy = 13.6×.5 =6.8And lambda = square root of 150 / V A° where V is the potential difference supplied to electron So lambda = square. Rt 150/6.8 that gives lambda = 4.71 A°
wavelength=12.27A°/√V So V=13.6×1.5×1.6×10^-19 Wavelength=4.7A° .........................................................
he energy given to electron is 13.6×1.5 = 20.4eV. So the kinetic energy of emitted electron is = 20.4-13.6 = 6.8eV. The wavelength of wave associated with electron moving with kinetic energy N eV is, (12.27) / (N^1/2) angstrom. 12.27/ (6.8)^1/2 = 4.7 angstrom.
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