Physical Chemistry> ATOMIC STRUCTURE...

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 ANGSTROMB)4.70 NANOMETERC)9.4 ANGSTROMD)9.40 NANOMETER

DEEPANKAR TULSIANI , 15 Years ago

Grade 12

7 Answers

askiitian.expert- chandra sekhar

Last Activity: 15 Years ago

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

nishchith

Last Activity: 9 Years ago

excess energy = 6.8 eV = 6.8 * 1.6 * 10^-19 J = 1.09 * 10^-18 J

h/mc = wavelength

here it can also be written as wavelength = h/sqrt(2mKE)

where h = 6.626 * 10^-34

m = 9.1 * 10^-31 kg

KE = eV = 1.09 * 10^-18 J

so substitute these answers and the right answer is 4.7 angstrom.

nishchith

Last Activity: 9 Years ago

excess energy = 6.8 eV = 6.8 * 1.6 * 10^-19 J = 1.09 * 10^-18 J

h/mc = wavelength

here it can also be written as wavelength = h/sqrt(2mKE)

where h = 6.626 * 10^-34

m = 9.1 * 10^-31 kg

KE = eV = 1.09 * 10^-18 J

so substitute these answers and the right answer is 4.7 angstrom.

Heet Surana

Last Activity: 8 Years ago

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

Ankit

Last Activity: 7 Years ago

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°

Alankrita Singh

Last Activity: 6 Years ago

wavelength=12.27A°/√V

So V=13.6×1.5×1.6×10^-19

Wavelength=4.7A° .........................................................

ankit singh

Last Activity: 4 Years ago

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.