The vapours of Hg absorb some electrons accelerated by a potential difference of `4.5` volt as a result of it 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.

Options

A. `3.63xx10^(6)m^(-1)`
B. `5.93xx10^(6)m^(-1)`
C. `5.93xx10^(6)cm^(-1)`
D. `5.62xx10^(6)m^(-1)`

Answer

To solve the problem, we need to determine the wave number of the light emitted when electrons are accelerated through a potential difference of 4.5 volts and collide with mercury (Hg) atoms. Here’s a step-by-step solution:

### Step 1: Calculate the energy of the electrons
The energy (E) gained by an electron when it is accelerated through a potential difference (V) is given by the formula:
\[ E = e \times V \]
where:
- \( e \) is the charge of an electron, approximately \( 1.6 \times 10^{-19} \) coulombs,
- \( V \) is the potential difference, which is \( 4.5 \) volts.

**Calculation:**
\[ E = 1.6 \times 10^{-19} \, \text{C} \times 4.5 \, \text{V} = 7.2 \times 10^{-19} \, \text{J} \]

### Step 2: Relate energy to wave number
The energy of a photon can also be expressed in terms of its wave number (\( \bar{\nu} \)) using the formula:
\[ E = h c \bar{\nu} \]
where:
- \( h \) is Planck's constant, approximately \( 6.626 \times 10^{-34} \, \text{J s} \),
- \( c \) is the speed of light, approximately \( 3.0 \times 10^{8} \, \text{m/s} \).

We can rearrange this equation to solve for wave number:
\[ \bar{\nu} = \frac{E}{h c} \]

### Step 3: Substitute the values into the wave number equation
Now, substituting the values we have calculated and known:
\[ \bar{\nu} = \frac{7.2 \times 10^{-19} \, \text{J}}{(6.626 \times 10^{-34} \, \text{J s}) \times (3.0 \times 10^{8} \, \text{m/s})} \]

**Calculation:**
1. Calculate \( h c \):
\[ h c = 6.626 \times 10^{-34} \times 3.0 \times 10^{8} = 1.9878 \times 10^{-25} \, \text{J m} \]

2. Now, calculate the wave number:
\[ \bar{\nu} = \frac{7.2 \times 10^{-19}}{1.9878 \times 10^{-25}} \approx 3.63 \times 10^{6} \, \text{m}^{-1} \]

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