Question

The value of Planck's constant is 6.63×10^–34Js. The velocity of light is 3.0×10⁸ ms^–1. Which value is closest to the wavelength in nanometers of a quantum of light with a frequency of 8×10¹⁵ s^–1:

• A. 2×10^–25
• B. 5×10^–18
• C. 4×10¹
• D. 3 ×10⁷

Answer 1

The Correct Option is C

 To find the wavelength of a quantum of light given its frequency, we can use the relationship between wavelength \((\lambda)\), frequency \((f)\), and the speed of light \((c)\). The formula connecting these quantities is:

\(c = \lambda \times f\)

Rearranging the formula to solve for wavelength \((\lambda)\), we get:

\(\lambda = \frac{c}{f}\)

Given:

• The speed of light, \(c = 3.0 \times 10^8\) m/s
• The frequency, \(f = 8 \times 10^{15}\) s^–1

 

Substitute the given values into the equation for wavelength:

\(\lambda = \frac{3.0 \times 10^8}{8 \times 10^{15}}\)

Calculating the above gives:

\(\lambda = \frac{3.0}{8} \times 10^{-7} \text{ m}\)

Simplifying \(\frac{3.0}{8}\), we get \(0.375 \times 10^{-7} \text{ m}\)

To convert meters to nanometers, we use the fact that \(1 \text{ m} = 10^9 \text{ nm}\). Thus:

\(\lambda = 0.375 \times 10^{-7} \times 10^9 \text{ nm}\)

Simplifying the above expression gives:

\(\lambda = 0.375 \times 10^2 \text{ nm}\)

\(\lambda = 37.5 \text{ nm}\)

The value closest to \(37.5\) nm from the given options is \(4 \times 10^1\) nm, which is equivalent to 40 nm.

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