Grade 12Physical Chemistry
7 Answers
askiitian.expert- chandra sekhar
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
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
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
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
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
wavelength=12.27A°/√V
So V=13.6×1.5×1.6×10^-19
Wavelength=4.7A° .........................................................
ankit singh
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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