The vapours of Hg absord some electron accelerated by a potiential diff. of 4.5 volt as a result of which light is emitted . If the full energy of single incident `e^(-)` is supposed to be converted into light emitted by single Hg atom , find the wave no. of the light

Explanation

To solve the problem, we need to find the wave number of the light emitted when the vapors of mercury (Hg) absorb energy from electrons accelerated by a potential difference of 4.5 volts.

### Step-by-Step Solution:

1. **Understanding the Energy of the Electron:**
The energy gained by an electron when it is accelerated through a potential difference (V) is given by the formula:
\[ E = eV \]
where \( e \) is the charge of the electron (approximately \( 1.6 \times 10^{-19} \) coulombs) and \( V \) is the potential difference in volts. Here, \( V = 4.5 \) volts.

2. **Calculating the Energy:**
Since we are given that the full energy of a single incident electron is converted into light emitted by a single Hg atom, we can directly use the potential difference:
\[ E = 4.5 \, \text{eV} \]

3. **Relating Energy to Wavelength:**
The energy of a photon (light) can also be expressed in terms of its wavelength (\( \lambda \)):
\[ E = \frac{hc}{\lambda} \]
where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{Js} \)) and \( c \) is the speed of light (\( 3 \times 10^8 \, \text{m/s} \)).

4. **Finding the Wavelength:**
Rearranging the equation for wavelength gives us:
\[ \lambda = \frac{hc}{E} \]
Substituting the values, we first convert the energy from eV to joules:
\[ E = 4.5 \, \text{eV} = 4.5 \times 1.6 \times 10^{-19} \, \text{J} = 7.2 \times 10^{-19} \, \text{J} \]

Now substituting \( h \), \( c \), and \( E \):
\[ \lambda = \frac{(6.626 \times 10^{-34} \, \text{Js})(3 \times 10^8 \, \text{m/s})}{7.2 \times 10^{-19} \, \text{J}} \]

5. **Calculating the Wavelength:**
Performing the calculation:
\[ \lambda = \frac{1.9878 \times 10^{-25}}{7.2 \times 10^{-19}} \approx 2.76 \times 10^{-7} \, \text{m} = 276 \, \text{nm} \]

6. **Finding the Wave Number:**
The wave number (\( \bar{\nu} \)) is defined as:
\[ \bar{\nu} = \frac{1}{\lambda} \]
Converting the wavelength to meters:
\[ \bar{\nu} = \frac{1}{2.76 \times 10^{-7}} \approx 3.63 \times 10^6 \, \text{m}^{-1} \]

### Final Answer:
The wave number of the light emitted is approximately:
\[ \bar{\nu} \approx 3.63 \times 10^6 \, \text{m}^{-1} \]