Search any question & find its solution
Question:
Answered & Verified by Expert
The value of Planck's constant is 6.63 $\times 10^{-34} \mathrm{Js}$. The velocity of light is $3.0 \times$ $10^8 \mathrm{~ms}^{-1}$. Which value is closest to the wavelength in nanometers of quantum of light with frequency of $8 \times 10^{15} \mathrm{~s}^{-1}$ ?
Options:
Solution:
1363 Upvotes
Verified Answer
The correct answer is:
$4 \times 10^1$
$\begin{aligned}
& c=v \times \lambda ; \lambda=\frac{c}{v} \\
& \Rightarrow \lambda=\frac{3 \times 10^8}{8 \times 10^{15}}=0.375 \times 10^{-7} \mathrm{~m} \\
& =0.375 \times 10^{-7} \times 10^9 \mathrm{~nm} \\
& =0.375 \times 10^2 \mathrm{~nm} \\
& =37.5 \mathrm{~nm} \\
& \approx 4 \times 10 \mathrm{~nm}
\end{aligned}$
& c=v \times \lambda ; \lambda=\frac{c}{v} \\
& \Rightarrow \lambda=\frac{3 \times 10^8}{8 \times 10^{15}}=0.375 \times 10^{-7} \mathrm{~m} \\
& =0.375 \times 10^{-7} \times 10^9 \mathrm{~nm} \\
& =0.375 \times 10^2 \mathrm{~nm} \\
& =37.5 \mathrm{~nm} \\
& \approx 4 \times 10 \mathrm{~nm}
\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.