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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

```
11 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
```
11 years ago
```							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.
```
5 years ago
```							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.
```
5 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
```
4 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°
```
2 years ago
```							 wavelength=12.27A°/√V So  V=13.6×1.5×1.6×10^-19 Wavelength=4.7A° .........................................................
```
2 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.
```
4 months ago
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