An electron having energy `20 e V` collides with a hydrogen atom in the ground state. As a result of the colllision , the atom is excite to a higher energy state and the electron is scattered with reduced velocity. The atom subsequentily returns to its ground state with emission of rediation of wavelength `1.216 xx 10^(-7) m`. Find the velocity of the scattered electron.

Answer

To solve the problem step by step, we will follow these steps:

### Step 1: Calculate the energy lost by the electron due to the emission of radiation.
The energy corresponding to the emitted radiation can be calculated using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
- \( h = 6.63 \times 10^{-34} \, \text{J s} \) (Planck's constant)
- \( c = 3 \times 10^{8} \, \text{m/s} \) (speed of light)
- \( \lambda = 1.216 \times 10^{-7} \, \text{m} \) (wavelength of emitted radiation)

Substituting the values:
\[ E = \frac{(6.63 \times 10^{-34}) \times (3 \times 10^{8})}{1.216 \times 10^{-7}} \]
Calculating this gives:
\[ E \approx 16.36 \times 10^{-19} \, \text{J} \]

### Step 2: Convert the initial energy of the electron from eV to Joules.
The initial energy of the electron is given as \( 20 \, \text{eV} \). To convert this to Joules, we use the conversion factor \( 1 \, \text{eV} = 1.6 \times 10^{-19} \, \text{J} \):
\[ E_{\text{initial}} = 20 \, \text{eV} \times (1.6 \times 10^{-19} \, \text{J/eV}) = 32 \times 10^{-19} \, \text{J} \]

### Step 3: Calculate the kinetic energy of the scattered electron.
The kinetic energy of the scattered electron can be found by subtracting the energy lost from the initial energy:
\[ E_{\text{scattered}} = E_{\text{initial}} - E_{\text{lost}} = (32 \times 10^{-19} \, \text{J}) - (16.36 \times 10^{-19} \, \text{J}) \]
Calculating this gives:
\[ E_{\text{scattered}} \approx 15.64 \times 10^{-19} \, \text{J} \]

### Step 4: Calculate the velocity of the scattered electron.
The kinetic energy of the scattered electron can also be expressed as:
\[ E = \frac{1}{2} mv^2 \]
where \( m \) is the mass of the electron, \( m = 9.11 \times 10^{-31} \, \text{kg} \). Rearranging for \( v \):
\[ v = \sqrt{\frac{2E}{m}} \]
Substituting the values:
\[ v = \sqrt{\frac{2 \times (15.64 \times 10^{-19})}{9.11 \times 10^{-31}}} \]
Calculating this gives:
\[ v \approx 1.86 \times 10^{6} \, \text{m/s} \]

### Final Answer:
The velocity of the scattered electron is approximately \( 1.86 \times 10^{6} \, \text{m/s} \).

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