Question

The value of Planck’s constant (h) is $6.63 \times {10^{ - 34}}Js$ . The velocity of light is $3.0 \times {10^8}m{s^{ - 1}}$ . Which value is closest to the wavelength (in meters) of a quantum of light with frequency of $8 \times {10^{15}}{s^{ - 1}}$ ?
 A.$3 \times {10^7}$
 B.$2 \times {10^{ - 25}}$
 C.$5 \times {10^{ - 18}}$
 D.$3.75 \times {10^{ - 8}}$

Answer 1

Hint: We know the velocity of light which is $3.0 \times {10^8}m{s^{ - 1}}$ and we have the quantum of light with frequency of $8 \times {10^{15}}{s^{ - 1}}$ . So, we can find the value of $\lambda $ by putting the value of c and $\vartheta $ in the equation, $\vartheta = \dfrac{C}{\lambda }$ .

Complete step by step answer:
We know that,
 $\vartheta = \dfrac{C}{\lambda }$
so, $\lambda = \dfrac{C}{\vartheta }$
Where,
 $\lambda $ = wavelength of the quantum of light
 $\vartheta $ = frequency of a quantum of light
c = speed of light
Also, h = $6.62 \times {10^{ - 34}}Js$ (given)
 $c = 3 \times {10^8}m{s^{ - 1}}$ (given)
 \[\lambda = 8 \times {10^{15}}{s^{ - 1}}\] (given)
Putting the value of c and $\vartheta $ in the equation, $\lambda = \dfrac{C}{\vartheta }$
 $\therefore \lambda = \dfrac{{3 \times {{10}^8}}}{{8 \times {{10}^{15}}}} = 0.375 \times {10^{ - 7}}$ $m = 3.75 \times {10^{ - 8}}$ m
So, $3.75 \times {10^{ - 8}}$ m is closest to the wavelength of a quantum of light with frequency of \[8 \times {10^{15}}{s^{ - 1}}\] .

Therefore, the correct answer is option (D).

Note: Wavelength of light is defined as the distance between the two successive crests or troughs of the light wave. Wavelength and frequency are very closely related to each other. Frequency is the number of waves that passes through a single given point in a specific amount of time. In order to know the wavelength of light, we must know the frequency and the speed of light. Frequency and wavelength are inversely proportional, which means longer the wavelength, lower the frequency. The frequency of the ray of light remains the same when it travels from one medium to the other medium.