Question

The value of Planck's constant is $ 6.63 \times 10^{-34} \,Js.$ The velocity of light is $3.0 \times 10^8\, ms^{-1}$. Which value is closest to the wavelength in nanometers of a quantum of light with frequency of $8 \times 10^{15}\, s^{-1}$?

• A. $4 \times 10^1$
• B. $3 \times 10^7$
• C. $2 \times 10^{25}$
• D. $5 \times 10^{-18}$

Answer 1

The Correct Option is A

To find the wavelength of a quantum of light, we can use the relationship between the speed of light (c), frequency (\nu), and wavelength (\lambda) given by the formula:

c = \lambda \nu

Where:

• c = 3.0 \times 10^8 \, \text{ms}^{-1} (speed of light)
• \nu = 8 \times 10^{15} \, \text{s}^{-1} (frequency)

We want to find \lambda, the wavelength.

Rearranging the formula to solve for \lambda gives:

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

Substituting the known values:

\lambda = \frac{3.0 \times 10^8}{8 \times 10^{15}} \, \text{m}

Calculating the above expression:

\lambda = \frac{3.0}{8} \times 10^{8-15} \, \text{m} = 0.375 \times 10^{-7} \, \text{m}

Converting meters to nanometers (1 m = 10^9 nm):

\lambda = 0.375 \times 10^{-7} \times 10^9 \, \text{nm} = 37.5 \, \text{nm}

This value is closest to the option 4 \times 10^1 \, \text{nm}.

Therefore, the correct answer is 4 \times 10^1.