The value of Planck's constant is `6.63 xx 10^-34 Js`. The velocity of light is `3 xx 10^8 m//sec`. Which value is closest to the wavelength of quantum of light with frequency of `8 xx 10^15 sec^-1` ?

Options

A. `5xx10^(-18)`
B. `4xx10^(1)`
C. `3xx10^(7)`
D. `2xx10^(-25)`

Answer

To find the wavelength of a quantum of light with a given frequency, we can use the formula that relates the speed of light (c), frequency (ν), and wavelength (λ):

\[ c = \nu \cdot \lambda \]

From this equation, we can rearrange it to solve for wavelength (λ):

\[ \lambda = \frac{c}{\nu} \]

### Step-by-Step Solution:

1. **Identify the given values**:
- Planck's constant (h) is not needed for this calculation.
- The speed of light (c) = \(3 \times 10^8 \, \text{m/s}\)
- The frequency (ν) = \(8 \times 10^{15} \, \text{s}^{-1}\)

2. **Substitute the values into the wavelength formula**:
\[ \lambda = \frac{3 \times 10^8 \, \text{m/s}}{8 \times 10^{15} \, \text{s}^{-1}} \]

3. **Perform the division**:
- First, divide the coefficients:
\[ \frac{3}{8} = 0.375 \]
- Next, divide the powers of ten:
\[ 10^8 / 10^{15} = 10^{-7} \]
- Combine these results:
\[ \lambda = 0.375 \times 10^{-7} \, \text{m} \]

4. **Convert to nanometers**:
- Since \(1 \, \text{m} = 10^9 \, \text{nm}\), we convert:
\[ \lambda = 0.375 \times 10^{-7} \, \text{m} \times 10^9 \, \text{nm/m} = 3.75 \times 10^{1} \, \text{nm} = 37.5 \, \text{nm} \]

5. **Round to the nearest significant figure**:
- Rounding \(37.5 \, \text{nm}\) gives approximately \(40 \, \text{nm}\).

### Final Answer:
The closest value to the wavelength of the quantum of light with a frequency of \(8 \times 10^{15} \, \text{s}^{-1}\) is approximately **40 nm**.