The circumference of the second Bohr orbit of electron in hydrogen atom is `600nm` . The potential difference that must be applied between the plates so that the electron have the de Broglie wavelength corresponding in this circumference is

Options

A. `10^(-5) V`
B. `(5)/(3)10^(-5) V`
C. `5 xx 10^(-5) V`
D. `3 xx10^(-5) V`

Explanation

To solve the problem step by step, we will follow the concepts of atomic physics, specifically the Bohr model and de Broglie wavelength.

### Step 1: Understand the given information
The circumference of the second Bohr orbit of the hydrogen atom is given as \(600 \, \text{nm}\). We need to find the potential difference \(V\) that must be applied so that the electron has a de Broglie wavelength corresponding to this circumference.

### Step 2: Relate the circumference to the de Broglie wavelength
The de Broglie wavelength \(\lambda\) of an electron in a circular orbit can be related to the circumference \(C\) of the orbit. For the second Bohr orbit, we have:
\[ C = 2 \pi r = n \lambda \]
where \(n\) is the principal quantum number. For the second Bohr orbit, \(n = 2\).

### Step 3: Calculate the de Broglie wavelength
Given that the circumference \(C = 600 \, \text{nm}\):
\[ \lambda = \frac{C}{n} = \frac{600 \, \text{nm}}{2} = 300 \, \text{nm} \]

### Step 4: Use the formula for the de Broglie wavelength of an electron
The de Broglie wavelength of an electron that has been accelerated through a potential difference \(V\) is given by:
\[ \lambda = \frac{h}{\sqrt{2meV}} \]
where:
- \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{Js}\)),
- \(m\) is the mass of the electron (\(9.11 \times 10^{-31} \, \text{kg}\)),
- \(e\) is the charge of the electron (\(1.6 \times 10^{-19} \, \text{C}\)),
- \(V\) is the potential difference.

### Step 5: Rearranging the formula to find \(V\)
We can rearrange the formula to solve for \(V\):
\[ V = \frac{h^2}{2m e \lambda^2} \]

### Step 6: Substitute the values
Substituting the known values into the equation:
- Convert \(\lambda\) from nanometers to meters: \(300 \, \text{nm} = 300 \times 10^{-9} \, \text{m}\).
\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (300 \times 10^{-9})^2} \]

### Step 7: Calculate \(V\)
Calculating the value:
\[ V = \frac{(6.626 \times 10^{-34})^2}{2 \times (9.11 \times 10^{-31}) \times (1.6 \times 10^{-19}) \times (9 \times 10^{-14})} \]
\[ V \approx \frac{4.39 \times 10^{-67}}{2.918 \times 10^{-48}} \approx 1.5 \times 10^{1} \, \text{V} \approx 15 \, \text{V} \]

### Final Answer
The potential difference that must be applied between the plates is approximately \(15 \, \text{V}\).

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