The value of planok,s constant is `6.63xx10^(-34)Js` the speed of light is` 3xx10^(17) nm s^(-1)` which value is closest to the wavelength in nanometer of a quantum of light with frequency of `6xx10^(15) S^(-1)`?

Options

A. 25
B. 50
C. 75
D. 10

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), wavelength (λ), and frequency (ν):

\[ C = \lambda \times \nu \]

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

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Speed of light (C) = \( 3 \times 10^{17} \) nm/s
- Frequency (ν) = \( 6 \times 10^{15} \) s\(^{-1}\)

2. **Substitute the values into the formula**:
\[ \lambda = \frac{3 \times 10^{17} \text{ nm/s}}{6 \times 10^{15} \text{ s}^{-1}} \]

3. **Perform the division**:
- First, divide the coefficients:
\[ \frac{3}{6} = 0.5 \]
- Next, divide the powers of ten:
\[ 10^{17} / 10^{15} = 10^{2} \]

4. **Combine the results**:
\[ \lambda = 0.5 \times 10^{2} \text{ nm} = 50 \text{ nm} \]

5. **Final result**:
The wavelength (λ) of the quantum of light is **50 nm**.